JEE Physics Electrostatic Potential and Capacitance MCQs Set 34

Answer: (d) zero

Explanation:
1. The electrostatic field is conservative.
2. Work done around any closed path is zero.

Teacher's Note:
1. Conservative forces do zero net work on a closed path.
2. Field lines are not closed loops.

Answer: (c) \( W_1=W_2=W_3 \)

Explanation:
1. Electrostatic work depends only on the end points, not on the path.

Teacher's Note:
1. Electrostatic force is conservative.
2. Work = change in potential energy.

Answer: (d) decreases then increases

Explanation:
1. Potential of two like charges is the sum \( V=kq(\frac1x+\frac1{d-x}) \).
2. It is large near each charge and has a minimum at the midpoint.
3. So it decreases, then increases.

Teacher's Note:
1. Near each charge potential is large.
2. Midpoint: field is zero, potential is minimum.

Answer: (b) a circle

Explanation:
1. Potential zero: \( \frac{kQ}{r_1}-\frac{2kQ}{r_2}=0 \), so \( r_2=2r_1 \).
2. The ratio of the distances from two fixed points is constant (not equal to 1): the locus is a circle (Apollonius circle).
3. A straight line would result only for equal and opposite charges.

Teacher's Note:
1. Constant ratio of distances gives a circle; ratio 1 gives the perpendicular bisector.
2. The charges have unequal magnitudes.

Answer: (a) 8 along negative x-axis

Explanation:
1. Field \( \vec E=-\nabla V \): \( E_x=-\frac{dV}{dx}=-8x \).
2. At \( x=1 \): \( E_x=-8 V/m \).
3. So the field is \( 8 V/m \) along the negative x-axis (\( E_y=E_z=0 \)).

Teacher's Note:
1. \( E \) = -dV/dx.
2. A negative sign means opposite to the \( +x \) direction.

Answer: (c) is zero along all the paths AB, AC, AD and AE

Explanation:
1. All points on the circle are at the same distance from the point charge, so they have the same potential.
2. Work done \( W=q\Delta V=0 \) for any path between them.

Teacher's Note:
1. Circle around a point charge is an equipotential.
2. Work = charge \( x \) potential difference.

Answer: (c) \( \frac{\sqrt2Q^2}{\pi\epsilon_0 a} \)

Explanation:
1. Distance from each corner to the centre is \( a/\sqrt2 \).
2. Potential at the centre \( V=4\cdot\frac{kQ}{a/\sqrt2}=\frac{4\sqrt2kQ}{a} \).
3. Work needed to remove \( -Q \) to \( \infty W=-Q(0-V)=QV=\frac{4\sqrt2kQ^2}{a}=\frac{\sqrt2Q^2}{\pi\epsilon_0a} \).

Teacher's Note:
1. Work by the external agent = \( q (V_{final} - V_{initial}) \).
2. k = 1/(4 pi epsilon-0).

Answer: (b) \( 2V/3 \)

Explanation:
1. Inside a conductor the potential is constant and equals the surface potential \( V \).
2. Outside, \( V\propto1/r \): at 15 cm \( V'=V\times\frac{10}{15}=\frac{2V}{3} \).

Teacher's Note:
1. Potential is constant inside the conductor.
2. Outside it behaves like a point charge at the centre.

Answer: (b) \( V\neq0\text{ and }E=0 \)

Explanation:
1. Potential is a scalar: all ten charges are at distance \( R \), so \( V=10k(-e)/R\neq0 \).
2. Fields from diametrically opposite electrons cancel by symmetry: \( E=0 \).

Teacher's Note:
1. Potential adds as scalars; fields add as vectors.
2. Symmetry cancels the field at the centre.

Answer: (a) \( Q(ae_1+be_2) \)

Explanation:
1. Work \( W=Q\vec E\cdot\vec r \).
2. Compute: \( W=Q(e_1a+e_2b) \) (the \( k \) component does not contribute).

Teacher's Note:
1. Work is the dot product of force and displacement.
2. The z-component of \( E \) gives no work as displacement has no z-part.

Answer: (b) \( \frac{-2q}{2+\sqrt2} \)

Explanation:
1. Energy \( U=k\left[\frac{q^2}{a}+\frac{Qq}{a}+\frac{Qq}{\sqrt2a}\right]=0 \).
2. So \( q+Q\left(1+\frac1{\sqrt2}\right)=0 \).
3. Hence \( Q=\frac{-q}{1+1/\sqrt2}=\frac{-\sqrt2q}{\sqrt2+1}=\frac{-2q}{2+\sqrt2} \).

Teacher's Note:
1. Sum the three pair energies.
2. Hypotenuse = \( a \sqrt{2} \).

Answer: (c) 0.1

Explanation:
1. Potential difference \( V=W/q=2/20=0.1 V \).
2. The distance is not needed.

Teacher's Note:
1. \( V = W/q \).
2. The path length is irrelevant.

Answer: (d) \( \frac{2Q}{4\pi\epsilon_0R}+\frac{q}{4\pi\epsilon_0R} \)

Explanation:
1. Potential at \( R/2 \) due to \( Q \) at the centre: \( \frac{kQ}{R/2}=\frac{2kQ}{R} \).
2. Potential inside a charged shell equals its surface potential: \( kq/R \).
3. Total: \( \frac{2Q}{4\pi\epsilon_0R}+\frac{q}{4\pi\epsilon_0R} \).

Teacher's Note:
1. Inside a shell the potential is constant = kq/R.
2. Add the two contributions as scalars.

Answer: (b) conductor

Explanation:
1. A conductor in equilibrium is an equipotential body.
2. The earth is a huge conductor, so its potential hardly changes when charge is added or removed, making it a convenient reference.

Teacher's Note:
1. Reference potential: earth or \( \infty \).
2. Huge conductor: potential practically constant.

Answer: (a) planes parallel to the yz-plane

Explanation:
1. Equipotential surfaces are perpendicular to the field lines.
2. The field is along \( x \), so the equipotentials are planes perpendicular to \( x \), i.e. parallel to the yz-plane (the field varies with \( x \), so they are unevenly spaced).

Teacher's Note:
1. Field is perpendicular to equipotentials.
2. Planes \( x \) = constant are parallel to the yz-plane.