Electric Charges and Fields
NCERT Class 12 Physics - Chapter 1 | High-Yield Advanced Assessment
Question 1
An uncharged, solid, isolated conducting sphere is placed in a uniform external electric field E0. Due to induction, a net charge separation occurs. If the potential at the center of the sphere is V0, what is the net electric potential at any arbitrary point inside the bulk of the conductor?
Concept Involved: Electrostatic properties of conductors (NCERT Section 1.5).
Explanation: In electrostatics, the electric field inside the bulk of a conductor is identically zero. The relation between electric field and potential gradient is given by E = -dV/dr. Since E = 0 everywhere inside the conductor, dV/dr = 0, meaning the electric potential V must be constant throughout the entire volume of the conductor, including its surface. Because the conductor forms an equipotential volume, the potential at any point inside is exactly equal to the potential at its center (V0). Options B and D are incorrect as potential does not vary with internal position. Option C is incorrect because a zero field implies a uniform potential, not necessarily zero potential.
Explanation: In electrostatics, the electric field inside the bulk of a conductor is identically zero. The relation between electric field and potential gradient is given by E = -dV/dr. Since E = 0 everywhere inside the conductor, dV/dr = 0, meaning the electric potential V must be constant throughout the entire volume of the conductor, including its surface. Because the conductor forms an equipotential volume, the potential at any point inside is exactly equal to the potential at its center (V0). Options B and D are incorrect as potential does not vary with internal position. Option C is incorrect because a zero field implies a uniform potential, not necessarily zero potential.
Question 2
Two identical small spheres carrying equal charges are suspended by insulating strings of equal length from a common point. The strings make an angle θ with each other in vacuum. The entire setup is then immersed in a liquid of density ρ and dielectric constant K. If the angle θ between the strings remains unchanged, what must be the density σ of the material of the spheres?
Concept Involved: Coulomb's Law in mediums and mechanical equilibrium (NCERT Section 1.6).
Explanation: In vacuum, the equilibrium condition gives tan(θ/2) = Fe / mg, where Fe is the electrostatic force and mg is the weight. When immersed in the liquid, the electrostatic force decreases to Fe' = Fe / K due to the dielectric medium. Simultaneously, the effective weight reduces due to buoyancy: mg' = mg - upthrust = V·σ·g - V·ρ·g = mg(1 - ρ/σ). For the angle to remain unchanged, tan(θ/2) must remain the same, so Fe / mg = Fe' / mg'. Substituting Fe' and mg' yields: Fe / mg = (Fe / K) / [mg(1 - ρ/σ)]. Simplifying gives 1 = 1 / [K(1 - ρ/σ)], which leads to 1 - ρ/σ = 1/K, or ρ/σ = 1 - 1/K = (K - 1)/K. Solving for σ, we find σ = K · ρ / (K - 1).
Explanation: In vacuum, the equilibrium condition gives tan(θ/2) = Fe / mg, where Fe is the electrostatic force and mg is the weight. When immersed in the liquid, the electrostatic force decreases to Fe' = Fe / K due to the dielectric medium. Simultaneously, the effective weight reduces due to buoyancy: mg' = mg - upthrust = V·σ·g - V·ρ·g = mg(1 - ρ/σ). For the angle to remain unchanged, tan(θ/2) must remain the same, so Fe / mg = Fe' / mg'. Substituting Fe' and mg' yields: Fe / mg = (Fe / K) / [mg(1 - ρ/σ)]. Simplifying gives 1 = 1 / [K(1 - ρ/σ)], which leads to 1 - ρ/σ = 1/K, or ρ/σ = 1 - 1/K = (K - 1)/K. Solving for σ, we find σ = K · ρ / (K - 1).
Question 3
Two point charges +q and +4q are fixed at a distance 'L' apart on a horizontal line. A third point charge Q is placed on the line joining them such that the entire system of three charges is in electrostatic equilibrium. What is the magnitude, sign, and position of the charge Q?
Concept Involved: Superposition Principle and System Equilibrium (NCERT Section 1.7).
Explanation: For the entire system to be in equilibrium, the net force on each individual charge must be zero. Let Q be placed at a distance x from +q. For the net force on Q to be zero: q·Q / (4πε0·x^2) = 4q·Q / [4πε0·(L - x)^2], which simplifies to 1/x = 2/(L - x) or L - x = 2x, giving x = L/3. To find the magnitude and sign of Q, consider the equilibrium of the charge +q: the repulsive force from +4q must balance the force from Q. Thus, q·4q / (4πε0·L^2) + q·Q / (4πε0·x^2) = 0. Substituting x = L/3, we get 4q/L^2 = -Q / (L/3)^2 = -9Q/L^2. Solving for Q yields Q = -4q/9. It must be negative to provide an attractive force counteracting the repulsion between the two positive charges.
Explanation: For the entire system to be in equilibrium, the net force on each individual charge must be zero. Let Q be placed at a distance x from +q. For the net force on Q to be zero: q·Q / (4πε0·x^2) = 4q·Q / [4πε0·(L - x)^2], which simplifies to 1/x = 2/(L - x) or L - x = 2x, giving x = L/3. To find the magnitude and sign of Q, consider the equilibrium of the charge +q: the repulsive force from +4q must balance the force from Q. Thus, q·4q / (4πε0·L^2) + q·Q / (4πε0·x^2) = 0. Substituting x = L/3, we get 4q/L^2 = -Q / (L/3)^2 = -9Q/L^2. Solving for Q yields Q = -4q/9. It must be negative to provide an attractive force counteracting the repulsion between the two positive charges.
Question 4
A thin, semi-circular ring of radius R is uniformly charged with a total positive charge Q. What is the magnitude and direction of the electric field intensity at the center of curvature of this semi-circular arc?
Concept Involved: Continuous charge distribution and electric field integration (NCERT Section 1.9).
Explanation: Let the linear charge density be λ = Q / (πR). Consider a small charge element dq = λ·dl = λ·R·dθ at an angle θ. The electric field due to this element at the center is dE = dq / (4πε0R^2) = λ·dθ / (4πε0R). Resolving dE into components along the axis of symmetry (dE·cosθ) and perpendicular to it (dE·sinθ), the perpendicular components cancel out across the symmetric halves. Integrating the axial components from -π/2 to +π/2 gives E = ∫ dE·cosθ = (λ / 4πε0R) · [sinθ] from -π/2 to +π/2 = 2λ / (4πε0R) = λ / (2πε0R). Substituting λ = Q / (πR), we get E = Q / (2π^2ε0R^2). The direction is away from the positively charged arc along its symmetry axis.
Explanation: Let the linear charge density be λ = Q / (πR). Consider a small charge element dq = λ·dl = λ·R·dθ at an angle θ. The electric field due to this element at the center is dE = dq / (4πε0R^2) = λ·dθ / (4πε0R). Resolving dE into components along the axis of symmetry (dE·cosθ) and perpendicular to it (dE·sinθ), the perpendicular components cancel out across the symmetric halves. Integrating the axial components from -π/2 to +π/2 gives E = ∫ dE·cosθ = (λ / 4πε0R) · [sinθ] from -π/2 to +π/2 = 2λ / (4πε0R) = λ / (2πε0R). Substituting λ = Q / (πR), we get E = Q / (2π^2ε0R^2). The direction is away from the positively charged arc along its symmetry axis.
Question 5
An electric dipole consisting of charges +q and -q separated by a distance 2a is placed along the x-axis with its center at the origin. A small test charge is placed at a point P on the x-axis at a distance r from the origin (r >> a). If the dipole is now rotated by 90° in the xy-plane about the origin, by what factor does the magnitude of the electric field at point P change?
Concept Involved: Electric field of an electric dipole on axial and equatorial points (NCERT Section 1.10).
Explanation: Initially, the dipole lies along the x-axis, so point P (at distance r on the x-axis) is an axial point. The electric field at a far axial point (r >> a) is given by E_axial = 2p / (4πε0r^3), where p = 2aq. When the dipole is rotated by 90° in the xy-plane, it aligns along the y-axis. Point P on the x-axis now lies on the equatorial plane of the dipole. The electric field at a far equatorial point is given by E_equatorial = p / (4πε0r^3). Comparing the two expressions, E_equatorial = E_axial / 2. Therefore, the magnitude of the electric field at point P is halved.
Explanation: Initially, the dipole lies along the x-axis, so point P (at distance r on the x-axis) is an axial point. The electric field at a far axial point (r >> a) is given by E_axial = 2p / (4πε0r^3), where p = 2aq. When the dipole is rotated by 90° in the xy-plane, it aligns along the y-axis. Point P on the x-axis now lies on the equatorial plane of the dipole. The electric field at a far equatorial point is given by E_equatorial = p / (4πε0r^3). Comparing the two expressions, E_equatorial = E_axial / 2. Therefore, the magnitude of the electric field at point P is halved.
Question 6
An electric dipole with dipole moment vector p is placed in a non-uniform electric field whose magnitude increases steadily along the positive x-direction. If the dipole moment vector p is aligned perfectly parallel to the positive x-axis, what will the dipole experience?
Concept Involved: Dipole in a non-uniform electric field (NCERT Section 1.11).
Explanation: The torque experienced by a dipole is given by the vector product τ = p × E. Since the dipole moment p is parallel to the electric field E (angle θ = 0°), the torque τ = pE sin(0°) = 0. However, because the field is non-uniform and increases along the positive x-direction, the positive charge (+q) experiences a larger force in the positive x-direction than the force experienced by the negative charge (-q) in the negative x-direction. Mathematically, the net force F ≈ p · (dE/dx). Since dE/dx > 0 and p is along the positive x-axis, the net force is positive, pulling the dipole in the direction of the increasing field (positive x-direction).
Explanation: The torque experienced by a dipole is given by the vector product τ = p × E. Since the dipole moment p is parallel to the electric field E (angle θ = 0°), the torque τ = pE sin(0°) = 0. However, because the field is non-uniform and increases along the positive x-direction, the positive charge (+q) experiences a larger force in the positive x-direction than the force experienced by the negative charge (-q) in the negative x-direction. Mathematically, the net force F ≈ p · (dE/dx). Since dE/dx > 0 and p is along the positive x-axis, the net force is positive, pulling the dipole in the direction of the increasing field (positive x-direction).
Question 7
A closed surface S encloses an electric dipole of moment p. A second closed surface S2 surrounds S. What is the total net electric flux passing through the surface S and surface S2 respectively?
Concept Involved: Gauss's Law and electric flux (NCERT Section 1.13).
Explanation: According to Gauss's Law, the net electric flux Φ through any closed surface is equal to the Net Enclosed Charge (Q_enclosed) divided by ε0. An electric dipole consists of two equal and opposite charges, +q and -q. Therefore, the net total charge of a dipole is (+q) + (-q) = 0. Since the surface S completely encloses the dipole, its enclosed charge is zero, making its flux zero. The surface S2 surrounds S, meaning it also encloses exactly the same system with a net charge of zero. Hence, the flux through both surfaces is identically zero, completely independent of the shape, size, or orientation of the surfaces.
Explanation: According to Gauss's Law, the net electric flux Φ through any closed surface is equal to the Net Enclosed Charge (Q_enclosed) divided by ε0. An electric dipole consists of two equal and opposite charges, +q and -q. Therefore, the net total charge of a dipole is (+q) + (-q) = 0. Since the surface S completely encloses the dipole, its enclosed charge is zero, making its flux zero. The surface S2 surrounds S, meaning it also encloses exactly the same system with a net charge of zero. Hence, the flux through both surfaces is identically zero, completely independent of the shape, size, or orientation of the surfaces.
Question 8
A point charge +q is placed exactly at one of the corners of a solid, uncharged cube of edge length 'a'. What is the total electric flux linked through the three faces of the cube that meet at the corner where the charge is located?
Concept Involved: Gauss's Law, flux definition, and geometry-based symmetry (NCERT Section 1.13).
Explanation: Electric flux is defined by the surface integral of E·dA. For a point charge located directly on a corner, the electric field lines radiate symmetrically outward from that point. For the three faces meeting at that specific corner, the field lines run completely parallel to the planes of these faces. This means the electric field vector E is perpendicular to the area vector dA (since dA is normal to the surface) at every point on these three faces. Thus, E·dA = E·dA·cos(90°) = 0. Consequently, the flux through these three adjacent faces is exactly zero. (Note: The total flux through the remaining three opposite faces combined is q / 8ε0, making the flux through each of those individual opposite faces q / 24ε0).
Explanation: Electric flux is defined by the surface integral of E·dA. For a point charge located directly on a corner, the electric field lines radiate symmetrically outward from that point. For the three faces meeting at that specific corner, the field lines run completely parallel to the planes of these faces. This means the electric field vector E is perpendicular to the area vector dA (since dA is normal to the surface) at every point on these three faces. Thus, E·dA = E·dA·cos(90°) = 0. Consequently, the flux through these three adjacent faces is exactly zero. (Note: The total flux through the remaining three opposite faces combined is q / 8ε0, making the flux through each of those individual opposite faces q / 24ε0).
Question 9
An infinite line of charge produces an electric field of magnitude E at a perpendicular distance r. If the linear charge density is λ, and the distance is doubled while the linear charge density is halved, what will be the new electric field intensity?
Concept Involved: Field due to an infinite line charge using Gauss's Law (NCERT Section 1.14.1).
Explanation: The electric field at a perpendicular distance r from an infinitely long straight wire with uniform linear charge density λ is given by the formula: E = λ / (2πε0r). This shows that E is directly proportional to λ and inversely proportional to r (E ∝ λ/r). Let the new field be E'. Given that the new linear charge density λ' = λ/2 and the new distance r' = 2r, we substitute these values into the proportionality relation: E' ∝ (λ/2) / (2r) = (1/4) · (λ/r). Therefore, E' = E / 4. The electric field drops to one-fourth of its initial value.
Explanation: The electric field at a perpendicular distance r from an infinitely long straight wire with uniform linear charge density λ is given by the formula: E = λ / (2πε0r). This shows that E is directly proportional to λ and inversely proportional to r (E ∝ λ/r). Let the new field be E'. Given that the new linear charge density λ' = λ/2 and the new distance r' = 2r, we substitute these values into the proportionality relation: E' ∝ (λ/2) / (2r) = (1/4) · (λ/r). Therefore, E' = E / 4. The electric field drops to one-fourth of its initial value.
Question 10
Consider a thin spherical shell of radius R carrying a total charge Q uniformly distributed over its surface. Which statement correctly describes the variation of the electric field E with the distance r from the center of the shell?
Concept Involved: Electric field of a charged thin spherical shell (NCERT Section 1.14.3).
Explanation: According to Gauss's Law, for any point inside the thin spherical shell (r < R), a spherical Gaussian surface encloses no net charge (Q_enclosed = 0). Consequently, the electric field E inside the shell is identically zero. For points outside the shell (r ≥ R), the entire charge Q can be assumed to be concentrated at the center, giving a field E = Q / (4πε0r^2), which follows an inverse-square law (E ∝ 1/r^2). At the surface (r = R), the field jumps discontinuously from 0 to its maximum value Q / (4πε0R^2). Thus, option B is the only mathematically correct description.
Explanation: According to Gauss's Law, for any point inside the thin spherical shell (r < R), a spherical Gaussian surface encloses no net charge (Q_enclosed = 0). Consequently, the electric field E inside the shell is identically zero. For points outside the shell (r ≥ R), the entire charge Q can be assumed to be concentrated at the center, giving a field E = Q / (4πε0r^2), which follows an inverse-square law (E ∝ 1/r^2). At the surface (r = R), the field jumps discontinuously from 0 to its maximum value Q / (4πε0R^2). Thus, option B is the only mathematically correct description.
Question 11
Two infinite parallel non-conducting sheets carry surface charge densities +σ and -2σ respectively. What is the magnitude of the net electric field in the region located between the two sheets?
Concept Involved: Electric field due to an infinite plane sheet of charge (NCERT Section 1.14.2).
Explanation: The electric field due to a single infinite flat non-conducting sheet with surface charge density σ is independent of distance and is given by E = σ / (2ε0). In the region between a positive sheet (+σ) and a negative sheet (-2σ), the electric field vectors produced by both sheets point in the same direction: away from the positive sheet and toward the negative sheet. Therefore, the fields add constructively. The total field is E_net = E1 + E2 = [σ / (2ε0)] + [2σ / (2ε0)] = 3σ / (2ε0).
Explanation: The electric field due to a single infinite flat non-conducting sheet with surface charge density σ is independent of distance and is given by E = σ / (2ε0). In the region between a positive sheet (+σ) and a negative sheet (-2σ), the electric field vectors produced by both sheets point in the same direction: away from the positive sheet and toward the negative sheet. Therefore, the fields add constructively. The total field is E_net = E1 + E2 = [σ / (2ε0)] + [2σ / (2ε0)] = 3σ / (2ε0).
Question 12
A small oil drop of mass m and carrying a net charge -q is balanced stationary in mid-air against gravity between two horizontal parallel plates. When a uniform vertical electric field E is active, it remains at rest. If the polarity of the electric field is suddenly reversed without changing its magnitude, what will be the instantaneous downward acceleration of the oil drop? (Neglect air buoyancy and viscous drag).
Concept Involved: Electric force balancing gravitational force (NCERT Section 1.8).
Explanation: Initially, the drop is in equilibrium, meaning the upward electric force balances the downward force of gravity. Since the charge is negative, the electric field must point downward to create an upward electrostatic force: qE = mg. When the field polarity is reversed, the electric field points upward, causing the electrostatic force on the negative charge to act downward with a magnitude of qE. Now, both gravity and the electric force act downward. The net downward force becomes F_net = mg + qE. Since qE = mg, F_net = mg + mg = 2mg. Using Newton's second law, the downward acceleration is a = F_net / m = 2mg / m = 2g.
Explanation: Initially, the drop is in equilibrium, meaning the upward electric force balances the downward force of gravity. Since the charge is negative, the electric field must point downward to create an upward electrostatic force: qE = mg. When the field polarity is reversed, the electric field points upward, causing the electrostatic force on the negative charge to act downward with a magnitude of qE. Now, both gravity and the electric force act downward. The net downward force becomes F_net = mg + qE. Since qE = mg, F_net = mg + mg = 2mg. Using Newton's second law, the downward acceleration is a = F_net / m = 2mg / m = 2g.
Question 13
Which of the following statements represents a strictly correct physical property of electrostatic field lines?
Concept Involved: Properties of Electric Field Lines (NCERT Section 1.10).
Explanation: Option C is a fundamental rule: field lines are always normal to the surface of a conductor. If they were not perpendicular, there would be a tangential component of the electric field along the surface, which would cause surface charges to move, violating the static equilibrium condition. Option A is false because electrostatic fields are conservative and cannot form closed loops. Option B is false because field lines can never intersect; if they did, it would imply two different directions for the net electric field at the intersection point, which is physically impossible. Option D is false because the relative density of lines is directly proportional to the field strength.
Explanation: Option C is a fundamental rule: field lines are always normal to the surface of a conductor. If they were not perpendicular, there would be a tangential component of the electric field along the surface, which would cause surface charges to move, violating the static equilibrium condition. Option A is false because electrostatic fields are conservative and cannot form closed loops. Option B is false because field lines can never intersect; if they did, it would imply two different directions for the net electric field at the intersection point, which is physically impossible. Option D is false because the relative density of lines is directly proportional to the field strength.
Question 14
An electric dipole with dipole moment p is aligned in a stable equilibrium configuration inside a uniform external electric field E. What is the total mechanical work required to rotate this dipole from its stable equilibrium position to its unstable equilibrium position?
Concept Involved: Torque and potential energy of a dipole in a uniform field (NCERT Section 1.11).
Explanation: Stable equilibrium occurs when the dipole moment p is aligned parallel to the field E (θ1 = 0°). Unstable equilibrium occurs when p is aligned anti-parallel to E (θ2 = 180°). The work done by an external agent in rotating a dipole is given by W = pE(cosθ1 - cosθ2). Substituting the angles for stable and unstable equilibrium, we get W = pE(cos 0° - cos 180°) = pE(1 - (-1)) = pE(1 + 1) = 2pE. Thus, option B is correct.
Explanation: Stable equilibrium occurs when the dipole moment p is aligned parallel to the field E (θ1 = 0°). Unstable equilibrium occurs when p is aligned anti-parallel to E (θ2 = 180°). The work done by an external agent in rotating a dipole is given by W = pE(cosθ1 - cosθ2). Substituting the angles for stable and unstable equilibrium, we get W = pE(cos 0° - cos 180°) = pE(1 - (-1)) = pE(1 + 1) = 2pE. Thus, option B is correct.
Question 15
A point charge q is suspended exactly at the center of an uncharged, hollow, thick spherical conducting shell of inner radius R1 and outer radius R2. What are the net induced surface charge densities (σ) on the inner and outer surfaces of the shell?
Concept Involved: Charge induction in conducting shells (NCERT Section 1.5).
Explanation: When a positive charge +q is placed at the center of the cavity, it induces an equal and opposite charge -q on the inner surface of the conductor to ensure that the electric field inside the metallic bulk remains zero. Since the conducting shell is uncharged and isolated, charge conservation requires an equal positive charge +q to be induced on its outer surface. Surface charge density is defined as charge divided by surface area (σ = Q / A). For the inner surface of radius R1, the area is 4πR1^2, so σ_inner = -q / (4πR1^2). For the outer surface of radius R2, the area is 4πR2^2, so σ_outer = +q / (4πR2^2).
Explanation: When a positive charge +q is placed at the center of the cavity, it induces an equal and opposite charge -q on the inner surface of the conductor to ensure that the electric field inside the metallic bulk remains zero. Since the conducting shell is uncharged and isolated, charge conservation requires an equal positive charge +q to be induced on its outer surface. Surface charge density is defined as charge divided by surface area (σ = Q / A). For the inner surface of radius R1, the area is 4πR1^2, so σ_inner = -q / (4πR1^2). For the outer surface of radius R2, the area is 4πR2^2, so σ_outer = +q / (4πR2^2).
Question 16
According to the theory of relativity, the mass of a particle increases with its velocity. If a highly charged elementary particle is accelerated to a relativistic velocity v = 0.9c (where c is the speed of light), how do its net electrical charge and its specific charge (charge-to-mass ratio) change?
Concept Involved: Invariance of charge vs variance of mass (NCERT Section 1.4).
Explanation: Electric charge is relativistically invariant, meaning the net charge of a particle is completely independent of its speed. However, relativistic mechanics states that the mass of a particle increases with speed according to the relation m = m0 / √(1 - v^2/c^2). The specific charge is defined as the charge-to-mass ratio (q/m). Since the numerator q remains constant while the denominator m increases at high velocities, the ratio q/m must decrease. This makes option C correct.
Explanation: Electric charge is relativistically invariant, meaning the net charge of a particle is completely independent of its speed. However, relativistic mechanics states that the mass of a particle increases with speed according to the relation m = m0 / √(1 - v^2/c^2). The specific charge is defined as the charge-to-mass ratio (q/m). Since the numerator q remains constant while the denominator m increases at high velocities, the ratio q/m must decrease. This makes option C correct.
Question 17
Five identical point charges, each of magnitude +q, are placed at five vertices of a regular hexagon of side length 'a'. What is the magnitude of the net electric field at the center of the hexagon?
Concept Involved: Superposition Principle and geometric symmetry (NCERT Section 1.7).
Explanation: In a perfectly symmetrical regular polygon with identical charges at all its vertices, the net electric field at the center is zero because the field vectors from opposite vertices cancel out. If we have five identical charges +q at five vertices of a regular hexagon, we can conceptually model this system as a complete hexagon with charges at all six vertices, combined with an additional virtual charge of -q at the vacant sixth vertex. Since the six symmetric charges add up to a net field of zero, the net field at the center is solely due to the virtual -q charge at the sixth vertex. The distance from any vertex to the center of a regular hexagon is equal to its side length 'a'. Thus, the magnitude of the field is simply E = q / (4πε0a^2).
Explanation: In a perfectly symmetrical regular polygon with identical charges at all its vertices, the net electric field at the center is zero because the field vectors from opposite vertices cancel out. If we have five identical charges +q at five vertices of a regular hexagon, we can conceptually model this system as a complete hexagon with charges at all six vertices, combined with an additional virtual charge of -q at the vacant sixth vertex. Since the six symmetric charges add up to a net field of zero, the net field at the center is solely due to the virtual -q charge at the sixth vertex. The distance from any vertex to the center of a regular hexagon is equal to its side length 'a'. Thus, the magnitude of the field is simply E = q / (4πε0a^2).
Question 18
A point charge +q is located at a perpendicular height 'd' directly above the geometric center of a flat, horizontal circular disc of radius R. What is the total electric flux passing through this disc?
Concept Involved: Integration of electric flux / solid angle concept (NCERT Section 1.12).
Explanation: The flux through an open circular disc subtended by a point charge can be calculated using integration or the concept of solid angle. The solid angle subtended by a disc of radius R at a point at distance d is given by Ω = 2π(1 - cosα), where cosα = d / √(R^2 + d^2). The total flux associated with a full sphere (solid angle 4π) is q/ε0. Therefore, the flux through the solid angle Ω is Φ = (q / ε0) · (Ω / 4π) = (q / ε0) · [2π(1 - cosα) / 4π] = (q / 2ε0) · (1 - cos&alpha忠). Substituting cosα, we get Φ = (q / 2ε0) · [1 - d / √(R^2 + d^2)].
Explanation: The flux through an open circular disc subtended by a point charge can be calculated using integration or the concept of solid angle. The solid angle subtended by a disc of radius R at a point at distance d is given by Ω = 2π(1 - cosα), where cosα = d / √(R^2 + d^2). The total flux associated with a full sphere (solid angle 4π) is q/ε0. Therefore, the flux through the solid angle Ω is Φ = (q / ε0) · (Ω / 4π) = (q / ε0) · [2π(1 - cosα) / 4π] = (q / 2ε0) · (1 - cos&alpha忠). Substituting cosα, we get Φ = (q / 2ε0) · [1 - d / √(R^2 + d^2)].
Question 19
A solid, non-conducting insulating sphere of radius R is charged non-uniformly such that its volume charge density varies with the radial distance r from the center as ρ(r) = ρ0 · (r / R), where ρ0 is a constant. What is the magnitude of the electric field inside the sphere at a distance r (r < R)?
Concept Involved: Gauss's Law applied to variable volume charge densities (NCERT Section 1.14).
Explanation: To find the electric field at an internal radius r, construct a concentric spherical Gaussian surface of radius r. The enclosed charge Q_enclosed is found by integrating the charge density over the volume: Q_enclosed = ∫ ρ(r) dV = ∫ [ρ0(x / R)] · (4πx^2 dx) from 0 to r = (4πρ0 / R) · [r^4 / 4] = πρ0r^4 / R. According to Gauss's Law, the flux is E · (4πr^2) = Q_enclosed / ε0. Substituting the enclosed charge gives E · (4πr^2) = (πρ0r^4) / (ε0R). Canceling 4π and r^2 from both sides yields E = ρ0r^2 / (4ε0R).
Explanation: To find the electric field at an internal radius r, construct a concentric spherical Gaussian surface of radius r. The enclosed charge Q_enclosed is found by integrating the charge density over the volume: Q_enclosed = ∫ ρ(r) dV = ∫ [ρ0(x / R)] · (4πx^2 dx) from 0 to r = (4πρ0 / R) · [r^4 / 4] = πρ0r^4 / R. According to Gauss's Law, the flux is E · (4πr^2) = Q_enclosed / ε0. Substituting the enclosed charge gives E · (4πr^2) = (πρ0r^4) / (ε0R). Canceling 4π and r^2 from both sides yields E = ρ0r^2 / (4ε0R).
Question 20
An electric dipole with dipole moment p and moment of inertia I is placed in a uniform external electric field E. If the dipole is slightly displaced from its stable equilibrium position by a small angle θ and released, it performs angular simple harmonic motion. What is its angular frequency (ω) of oscillation?
Concept Involved: Dynamics of an electric dipole and SHM (NCERT Section 1.11).
Explanation: When a dipole is rotated by an angle θ from its equilibrium position, it experiences a restoring torque given by τ = -pE sinθ. For small angular displacements, we can apply the approximation sinθ ≈ θ, which simplifies the restoring torque to τ ≈ -pEθ. The rotational analogue of Newton's second law is τ = I·α, where I is the moment of inertia and α is the angular acceleration. Equating the torque expressions gives I·α = -pEθ, which can be rewritten as α = -(pE / I)θ. This matches the standard equation for simple harmonic motion, α = -ω^2θ. By comparison, ω^2 = pE / I, which gives an angular frequency of ω = √(pE / I).
Explanation: When a dipole is rotated by an angle θ from its equilibrium position, it experiences a restoring torque given by τ = -pE sinθ. For small angular displacements, we can apply the approximation sinθ ≈ θ, which simplifies the restoring torque to τ ≈ -pEθ. The rotational analogue of Newton's second law is τ = I·α, where I is the moment of inertia and α is the angular acceleration. Equating the torque expressions gives I·α = -pEθ, which can be rewritten as α = -(pE / I)θ. This matches the standard equation for simple harmonic motion, α = -ω^2θ. By comparison, ω^2 = pE / I, which gives an angular frequency of ω = √(pE / I).
Question 21
A point charge +Q is fixed at the origin. Another point charge -q of mass m enters the neighborhood and moves in a circular orbit of radius R around +Q under the influence of the mutual electrostatic attraction. What is the time period of revolution T for this circular path?
Concept Involved: Coulomb's Law combined with centripetal force requirements (NCERT Section 1.6).
Explanation: The attractive electrostatic force between the two charges provides the necessary centripetal force keeping the charge -q in its circular orbit. Mathematically, F_electrostatic = F_centripetal, which translates to qQ / (4πε0R^2) = m·v^2 / R = m·ω^2·R. Solving for ω^2, we get ω^2 = qQ / (4πε0mR^3). Since the relation between the time period and angular velocity is T = 2π / ω, squaring both sides gives T^2 = 4π^2 / ω^2 = 4π^2 · (4πε0mR^3 / qQ) = 16π^3ε0mR^3 / qQ. Taking the square root of both sides yields T = 4πR · √(πε0mR / qQ).
Explanation: The attractive electrostatic force between the two charges provides the necessary centripetal force keeping the charge -q in its circular orbit. Mathematically, F_electrostatic = F_centripetal, which translates to qQ / (4πε0R^2) = m·v^2 / R = m·ω^2·R. Solving for ω^2, we get ω^2 = qQ / (4πε0mR^3). Since the relation between the time period and angular velocity is T = 2π / ω, squaring both sides gives T^2 = 4π^2 / ω^2 = 4π^2 · (4πε0mR^3 / qQ) = 16π^3ε0mR^3 / qQ. Taking the square root of both sides yields T = 4πR · √(πε0mR / qQ).
Question 22
The electric field in a certain region of space is given by the vector expression E = E0·x i, where E0 is a positive constant and i is the unit vector along the x-axis. What is the total net charge enclosed within an imaginary cube of side length 'a' bounded by the planes x = 0, x = a, y = 0, y = a, z = 0, and z = a?
Concept Involved: Flux evaluation in non-uniform fields using Gauss's Law (NCERT Section 1.13).
Explanation: The electric field E = E0·x i has only an x-component, meaning its vectors are parallel to the x-axis. Consequently, the electric flux through the four faces of the cube that run parallel to the x-axis (top, bottom, front, and back) is zero because the field vectors are parallel to the surfaces. This leaves only the two faces perpendicular to the x-axis: the face at x = 0 and the face at x = a. At x = 0, the electric field is E = E0·(0) = 0, so the flux entering this left face is zero. At x = a, the electric field is E = E0·a i, and the area vector points outward in the positive x-direction (dA = a^2 i). The flux exiting this right face is Φ = E · dA = (E0·a) · a^2 = E0a^3. According to Gauss's Law, Q_enclosed = ε0 · Φ_net = ε0E0a^3.
Explanation: The electric field E = E0·x i has only an x-component, meaning its vectors are parallel to the x-axis. Consequently, the electric flux through the four faces of the cube that run parallel to the x-axis (top, bottom, front, and back) is zero because the field vectors are parallel to the surfaces. This leaves only the two faces perpendicular to the x-axis: the face at x = 0 and the face at x = a. At x = 0, the electric field is E = E0·(0) = 0, so the flux entering this left face is zero. At x = a, the electric field is E = E0·a i, and the area vector points outward in the positive x-direction (dA = a^2 i). The flux exiting this right face is Φ = E · dA = (E0·a) · a^2 = E0a^3. According to Gauss's Law, Q_enclosed = ε0 · Φ_net = ε0E0a^3.
Question 23
A point charge q is placed just a tiny distance above the exact geometric center of a flat hemispherical open bowl surface. What is the net electric flux passing through the curved surface of this hemisphere?
Concept Involved: Closed surface symmetry and Gauss's Law (NCERT Section 1.13).
Explanation: To use Gauss's Law, we can mentally close the open hemispherical bowl by placing an identical inverted hemispherical bowl directly over it. This forms a complete, closed spherical surface enclosing the point charge q at its center. By Gauss's Law, the total flux passing through the entire closed sphere is q/ε0. Because the charge sits symmetrically at the center, the flux divides equally between the upper and lower halves. Therefore, the flux through the single open curved hemispherical surface is exactly half of the total flux: Φ = q / (2ε0).
Explanation: To use Gauss's Law, we can mentally close the open hemispherical bowl by placing an identical inverted hemispherical bowl directly over it. This forms a complete, closed spherical surface enclosing the point charge q at its center. By Gauss's Law, the total flux passing through the entire closed sphere is q/ε0. Because the charge sits symmetrically at the center, the flux divides equally between the upper and lower halves. Therefore, the flux through the single open curved hemispherical surface is exactly half of the total flux: Φ = q / (2ε0).
Question 24
An uncharged, insulated hollow conducting sphere has a point charge +q placed inside its internal cavity, but positioned off-center (eccentric position). Which statement correctly describes the resulting electric field lines and charge distributions?
Concept Involved: Electrostatic shielding and conductor behaviors (NCERT Section 1.5).
Explanation: Because the point charge +q is located off-center inside the cavity, it creates an asymmetric electric field that induces a non-uniform negative charge distribution on the inner cavity wall to ensure the field within the metal remains zero. However, the conducting bulk of the sphere shields the outer surface from the asymmetric interior. The induced positive charge on the outer surface spreads out uniformly across the sphere's exterior to minimize its electrostatic potential energy. As a result, the electric field lines exiting the outer surface are perfectly symmetric and point radially outward, making option B correct.
Explanation: Because the point charge +q is located off-center inside the cavity, it creates an asymmetric electric field that induces a non-uniform negative charge distribution on the inner cavity wall to ensure the field within the metal remains zero. However, the conducting bulk of the sphere shields the outer surface from the asymmetric interior. The induced positive charge on the outer surface spreads out uniformly across the sphere's exterior to minimize its electrostatic potential energy. As a result, the electric field lines exiting the outer surface are perfectly symmetric and point radially outward, making option B correct.
Question 25
A uniformly charged ring of radius R carries a total positive charge Q. At what distance x along the axis of the ring, measured from its geometric center, does the electric field intensity reach its maximum value?
Concept Involved: Maximization of axial field of a ring (NCERT Section 1.9).
Explanation: The electric field at a distance x along the axis of a uniformly charged ring of radius R is given by the formula: E(x) = Q·x / [4πε0 · (R^2 + x^2)^(3/2)]. To find the position where the field reaches its maximum, we take the derivative of E with respect to x and set it equal to zero (dE/dx = 0). Using the calculus quotient rule, we find that the derivative vanishes when R^2 - 2x^2 = 0, which simplifies to x^2 = R^2 / 2. Solving for x gives x = R / √2. At the center (x = 0), the field is zero, and at very large distances (x → ∞), the field approaches zero. The maximum field value occurs exactly at x = R / √2.
Explanation: The electric field at a distance x along the axis of a uniformly charged ring of radius R is given by the formula: E(x) = Q·x / [4πε0 · (R^2 + x^2)^(3/2)]. To find the position where the field reaches its maximum, we take the derivative of E with respect to x and set it equal to zero (dE/dx = 0). Using the calculus quotient rule, we find that the derivative vanishes when R^2 - 2x^2 = 0, which simplifies to x^2 = R^2 / 2. Solving for x gives x = R / √2. At the center (x = 0), the field is zero, and at very large distances (x → ∞), the field approaches zero. The maximum field value occurs exactly at x = R / √2.
Question 26
Two short electric dipoles with parallel dipole moments p1 and p2 are positioned along the same straight line (co-axial configuration) separated by a large distance r. What is the mathematical dependency of the mutual electrostatic force F interacting between them?
Concept Involved: Inter-dipole forces and field gradients (NCERT Section 1.10).
Explanation: The electric field produced by the first dipole p1 at a distance r along its axis is given by E = 2p1 / (4πε0r^3). The force experienced by a second dipole p2 placed in this field gradient is given by F = p2 · (dE/dr). Differentiating the electric field expression with respect to r gives dE/dr = -6p1 / (4πε0r^4). Substituting this derivative into the force equation yields F = -12p1p2 / (4πε0r^4). Ignoring the negative sign (which indicates the direction of the force), the magnitude of the force scales as F ∝ 1/r^4. This is a much faster drop-off than the standard 1/r^2 relation seen in Coulomb's Law for point charges.
Explanation: The electric field produced by the first dipole p1 at a distance r along its axis is given by E = 2p1 / (4πε0r^3). The force experienced by a second dipole p2 placed in this field gradient is given by F = p2 · (dE/dr). Differentiating the electric field expression with respect to r gives dE/dr = -6p1 / (4πε0r^4). Substituting this derivative into the force equation yields F = -12p1p2 / (4πε0r^4). Ignoring the negative sign (which indicates the direction of the force), the magnitude of the force scales as F ∝ 1/r^4. This is a much faster drop-off than the standard 1/r^2 relation seen in Coulomb's Law for point charges.
Question 27
An electron and a proton are released from rest in a uniform external electric field E. If we ignore their mutual gravitational attraction, how do their accelerations (a_e, a_p) and the times (t_e, t_p) required to traverse an identical distance d compare?
Concept Involved: Motion of charged particles in a uniform electric field (NCERT Section 1.8).
Explanation: Both the electron and proton carry an identical magnitude of elementary charge |e|, meaning they experience the same magnitude of electrostatic force F = |e|E in the uniform field. According to Newton's second law, acceleration is force divided by mass (a = F/m). Since the mass of the electron is roughly 1836 times smaller than the mass of the proton, the electron develops a much larger acceleration (a_e > a_p). The time required to cover a distance d starting from rest is given by kinematic equations: d = (1/2)·a·t^2, which simplifies to t = √(2d/a). Because the electron has a higher acceleration, it takes less time to cover the same distance (t_e < t_p).
Explanation: Both the electron and proton carry an identical magnitude of elementary charge |e|, meaning they experience the same magnitude of electrostatic force F = |e|E in the uniform field. According to Newton's second law, acceleration is force divided by mass (a = F/m). Since the mass of the electron is roughly 1836 times smaller than the mass of the proton, the electron develops a much larger acceleration (a_e > a_p). The time required to cover a distance d starting from rest is given by kinematic equations: d = (1/2)·a·t^2, which simplifies to t = √(2d/a). Because the electron has a higher acceleration, it takes less time to cover the same distance (t_e < t_p).
Question 28
A tiny positive test charge is released from rest in an electrostatic field configuration where the electric field lines are highly curved. Assuming no other forces act on the particle, will it track and move along the trajectory of the field line?
Concept Involved: Physical interpretation of electric field lines (NCERT Section 1.10).
Explanation: An electric field line is defined such that the tangent drawn at any point gives the direction of the electric field vector E, which matches the direction of the instantaneous force acting on a positive charge. When the charge is released from rest, its initial acceleration points tangent to the field line, starting its motion along it. However, as it gains velocity, its inertia carries it forward along its current velocity vector. For the particle to follow a curved path, a centripetal force is required. Since the electric force only acts tangent to the field line, it cannot provide the necessary perpendicular component to keep the particle on a curved trajectory. Therefore, the particle will deviate from the curved field line.
Explanation: An electric field line is defined such that the tangent drawn at any point gives the direction of the electric field vector E, which matches the direction of the instantaneous force acting on a positive charge. When the charge is released from rest, its initial acceleration points tangent to the field line, starting its motion along it. However, as it gains velocity, its inertia carries it forward along its current velocity vector. For the particle to follow a curved path, a centripetal force is required. Since the electric force only acts tangent to the field line, it cannot provide the necessary perpendicular component to keep the particle on a curved trajectory. Therefore, the particle will deviate from the curved field line.
Question 29
An infinite line of charge is surrounded by a coaxial, uncharged, thick conducting cylindrical pipe of inner radius R1 and outer radius R2. If the linear charge density of the central wire is λ, what is the electric field intensity in the region inside the bulk metal of the cylindrical pipe (R1 < r < R2)?
Concept Involved: Gauss's Law applied to conductors (NCERT Section 1.14.1 / 1.5).
Explanation: This question tests a fundamental principle: the electric field inside the bulk material of any conductor in electrostatic equilibrium is always zero. When the central wire carrying a linear charge density λ is placed inside the pipe, it induces a linear charge density of -λ on the inner surface (r = R1) and +λ on the outer surface (r = R2) of the conducting cylinder. If we construct a cylindrical Gaussian surface within the metal bulk (R1 < r < R2), the enclosed charge consists of the central line charge (+λ per unit length) and the induced inner surface charge (-\lambda per unit length), which sum to zero. Thus, the net enclosed charge is zero, and the electric field inside the conductor's bulk is exactly zero.
Explanation: This question tests a fundamental principle: the electric field inside the bulk material of any conductor in electrostatic equilibrium is always zero. When the central wire carrying a linear charge density λ is placed inside the pipe, it induces a linear charge density of -λ on the inner surface (r = R1) and +λ on the outer surface (r = R2) of the conducting cylinder. If we construct a cylindrical Gaussian surface within the metal bulk (R1 < r < R2), the enclosed charge consists of the central line charge (+λ per unit length) and the induced inner surface charge (-\lambda per unit length), which sum to zero. Thus, the net enclosed charge is zero, and the electric field inside the conductor's bulk is exactly zero.
Question 30
What is the total electrostatic potential energy stored in an isolated system of two identical point charges +q separated by a distance r in a vacuum, and what force would be needed to hold them in place?
Concept Involved: Coulomb's Law and Electrostatic Energy (NCERT Section 1.6).
Explanation: The work done by an external agent to bring two like charges from infinity to a separation distance r is stored as electrostatic potential energy, given by U = q^2 / (4πε0r). Because both charges are positive, they exert a mutual repulsive Coulomb force on each other given by F = q^2 / (4πε0r^2). To counteract this repulsion and hold the charges stationary, an external agent must apply an equal and opposite compressive holding force of magnitude q^2 / (4πε0r^2) directed inward. This satisfies both energy and force equilibrium conditions, confirming option B as correct.
Explanation: The work done by an external agent to bring two like charges from infinity to a separation distance r is stored as electrostatic potential energy, given by U = q^2 / (4πε0r). Because both charges are positive, they exert a mutual repulsive Coulomb force on each other given by F = q^2 / (4πε0r^2). To counteract this repulsion and hold the charges stationary, an external agent must apply an equal and opposite compressive holding force of magnitude q^2 / (4πε0r^2) directed inward. This satisfies both energy and force equilibrium conditions, confirming option B as correct.
