JEE Physics Electrostatic Potential and Capacitance MCQs Set 25

Answer: (b) \( E = \frac{Q_2}{4\pi\epsilon_0r_B^2},\; V = \frac{Q_1}{4\pi\epsilon_0R_1}+\frac{Q_2}{4\pi\epsilon_0r_B} \)

Explanation:
1. Field in the gap comes only from the inner charge: \( E = \frac{kQ_2}{r_B^2} \) (the outer shell gives zero inside).
2. Potential = potential of the outer shell (constant inside it) + potential due to the inner charge at \( r_B \).

Teacher's Note:
1. A uniformly charged shell gives no field inside but a constant potential equal to its surface value.
2. Add potentials as scalars.

Answer: (a) 8 J

Explanation:
1. For a point charge \( V = Er = 4\times2 = 8 \) V.
2. W = qV = 1 × 8 = 8 J.

Teacher's Note:
1. For a point charge, V = E r.
2. Work from infinity equals q times the potential.

Answer: (c) Only system (1)

Explanation:
1. System (1): \( U \propto \frac{(-1)(1)}{2}+\frac{(-1)(1)}{2}+\frac{(1)(1)}{1} = -\frac12-\frac12+1 = 0 \).
2. System (2): all terms are positive, so U > 0.
3. System (3): 2/1 − 2/2 − 1/3 is not zero.

Teacher's Note:
1. Potential energy of a system is the sum of kq₁q₂/r over all pairs.
2. Systems with only like charges cannot have zero energy.

Answer: (a) 10 V

Explanation:
1. Inside a conductor the field is zero, so the potential is constant and equals its surface value, 10 V.
2. At 5 cm outside the surface V = 10 × 5/10 = 5 V, which is different.

Teacher's Note:
1. Constant potential inside a conductor does not mean zero potential.
2. Outside, V falls as 1/r.

Answer: (d) \( \frac{q^2}{4\pi\epsilon_0a} \)

Explanation:
1. Initial energy: \( kq^2\left(-\frac1a-\frac1a+\frac1{2a}\right) = -\frac{3kq^2}{2a} \).
2. After the exchange the arrangement is −q, −q, +q: \( kq^2\left(\frac1a-\frac1{2a}-\frac1a\right) = -\frac{kq^2}{2a} \).
3. Work = change in energy \( = \frac{kq^2}{a} = \frac{q^2}{4\pi\epsilon_0a} \).

Teacher's Note:
1. Compute the energy of the configuration before and after.
2. Work done against the field equals the increase in potential energy.

Answer: (b) zero

Explanation:
1. Initial energy: \( \frac{k}{a}\left[-2q^2-2q^2+4q^2\right] = 0 \).
2. The final energy is the same expression divided by 2, also zero. So the work is zero.

Teacher's Note:
1. Work equals the change in energy.
2. Check the initial energy before assuming any work is needed.

Answer: (c) R/3

Explanation:
1. Centre: \( V_0 = \frac{3kQ}{2R} \). Half of it is \( \frac{3kQ}{4R} \).
2. Outside: \( \frac{kQ}{r} = \frac{3kQ}{4R} \Rightarrow r = \frac{4R}{3} \), i.e. R/3 from the surface.

Teacher's Note:
1. Potential at the centre of a uniformly charged sphere is 1.5 times the surface value.
2. Measure the distance from the surface, not from the centre.

Answer: (d) \( -\frac{Q}{\sqrt3} \)

Explanation:
1. Side a: the centroid is at \( a/\sqrt3 \) from each vertex.
2. \( U = \frac{3kQ^2}{a}+3\cdot\frac{kQq\sqrt3}{a} = 0 \Rightarrow q = -\frac{Q}{\sqrt3} \).

Teacher's Note:
1. Count the three vertex–vertex pairs and the three vertex–centre pairs.
2. Distance from the centroid to a vertex is a/√3.

Answer: (b) zero

Explanation:
1. The electric force is conservative and always perpendicular to the displacement on a circular path around the fixed charge.
2. So the work done is zero.

Teacher's Note:
1. Work done by a conservative force over a closed path is zero.
2. Force is radial, displacement is tangential.

Answer: (a) 10 V

Explanation:
1. \( E = F/q = 1000 \) N/C.
2. \( \Delta V = Ed = 1000\times0.01 = 10 \) V.

Teacher's Note:
1. Convert cm to m before multiplying.
2. Field in volts per metre equals force per unit charge.

Answer: (c) Region 2

Explanation:
1. \( E_x = -\frac{dV}{dx} \), so the field magnitude is largest where the slope is steepest.
2. The steepest slope is in region 2.

Teacher's Note:
1. Field is the negative gradient of potential.
2. A flat potential means zero field.

Answer: (a) \( 2k\sqrt6 \)

Explanation:
1. \( E_x = -4kx,\; E_y = 2ky,\; E_z = -2kz \).
2. At (1,1,1): \( |E| = k\sqrt{16+4+4} = 2k\sqrt6 \).

Teacher's Note:
1. Take partial derivatives with respect to each coordinate.
2. Magnitude is the root of the sum of squares.

Answer: (d) The potentials at A and B are equal and the potential at C is lower than at A

Explanation:
1. Equipotential surfaces are perpendicular to the field, so points on the same vertical line have equal potential.
2. Potential falls in the direction of the field, so C (to the right) is at a lower potential.

Teacher's Note:
1. Moving along the field lowers the potential.
2. Moving perpendicular to the field does not change it.

Answer: (b) flows from the inner shell to the outer shell

Explanation:
1. Before connection, \( V_{inner} = \frac{k\cdot3Q}{R}+\frac{kQ}{2R} \) is greater than \( V_{outer} = \frac{k\cdot4Q}{2R} \).
2. Positive charge flows from higher to lower potential: inner to outer.

Teacher's Note:
1. Charge always flows to equalise potential.
2. When joined, all charge ends on the outer surface.

Answer: (c) the distribution would depend on whether other charges are present or not

Explanation:
1. In a conductor the charge resides on the surface, not in the volume.
2. On an isolated sphere it is uniform, but nearby charges redistribute it, so uniformity is not guaranteed.

Teacher's Note:
1. Induced charges change the surface distribution.
2. Charge never sits inside a conductor in equilibrium.

Answer: (d) zero

Explanation:
1. Each dipole has zero net charge, so the total enclosed charge is zero.
2. By Gauss's law the net flux is zero.

Teacher's Note:
1. Flux depends on net enclosed charge.
2. Individual dipole fields do pass through the surface but cancel overall.

Answer: (b) the electric field outside Q is zero

Explanation:
1. Q is at zero potential and the field outside an earthed enclosing shell is zero: the negative charge induced on Q equals the charge of P.
2. The field between shells varies as 1/r², and the potential inside P is positive, not zero.

Teacher's Note:
1. An earthed outer shell screens the outside world from the inner charge.
2. The field between concentric shells is radial and not uniform.

Answer: (a) x = −2Q and x′ = 0

Explanation:
1. The total charge of the system is 2Q, so each outer surface carries Q.
2. Left plate: outer Q, inner 2Q. The middle plate's left face is −2Q (opposite to the inner face of the left plate), so its right face is 0 and the right plate's inner face is also 0.

Teacher's Note:
1. Outermost faces carry equal charge, half of the total.
2. Facing surfaces carry equal and opposite charges.

Answer: (c) 1 : 4

Explanation:
1. Volume conservation: \( R = 4r \), charge \( 64q \).
2. \( \sigma_{small} = \frac{q}{4\pi r^2},\; \sigma_{big} = \frac{64q}{4\pi(4r)^2} = \frac{4q}{4\pi r^2} \), so the ratio is 1 : 4.

Teacher's Note:
1. Radius of the big drop follows from conservation of volume.
2. Surface density = charge / surface area.

Answer: (a) a metallic surface

Explanation:
1. Inside a closed conducting enclosure the electrostatic field is zero (electrostatic shielding).
2. Insulators and imaginary Gaussian surfaces do not screen the field.

Teacher's Note:
1. Shielding needs a conductor.
2. A Gaussian surface is only a mathematical construct.