JEE Physics Electrostatic Potential and Capacitance MCQs Set 10

Answer: (a) zero

Explanation:
1. Induction only separates the charges on the sphere: the side near the point charge becomes negative and the far side becomes positive.
2. No charge enters or leaves the sphere, so the net charge stays zero.

Teacher's Note:
1. Induced charges are equal and opposite. A neutral conductor remains neutral overall, although it is attracted by the nearby charge.

Answer: (a) 623.5 N

Explanation:
1. Force between two charges: \( F = \dfrac{9\times10^9\times(10^{-5})^2}{(0.05)^2} = 360 \) N.
2. Each charge feels two such forces at 60°: \( F_{net} = 2F\cos30^\circ = \sqrt3\times360 \approx 623.5 \) N.

Teacher's Note:
1. Add the two forces as vectors. The angle between them is 60°, so the resultant is √3 times one force.

Answer: (b) \( \dfrac{4Q_1}{Q_2} \)

Explanation:
1. \( F_1 = \dfrac{kQ_1Q_2}{d^2} \).
2. After contact each sphere has \( \dfrac{Q_1+Q_2}{2} \approx \dfrac{Q_2}{2} \), so \( F_2 = \dfrac{k(Q_2/2)^2}{d^2} = \dfrac{kQ_2^2}{4d^2} \).
3. \( \dfrac{F_1}{F_2} = \dfrac{4Q_1}{Q_2} \).

Teacher's Note:
1. On contact, identical conductors share the total charge equally. Use Q2 + Q1 ≈ Q2 because Q2 is much larger.

Answer: (b) \( -\dfrac{Q}{4} \)

Explanation:
1. Take the charges Q a distance d apart. For one of the charges Q to be in equilibrium: \( \dfrac{kQq}{(d/2)^2} + \dfrac{kQ^2}{d^2} = 0 \) (the forces are opposite).
2. \( 4q = -Q \Rightarrow q = -\dfrac{Q}{4} \).

Teacher's Note:
1. Equilibrium of the whole system means each charge is in equilibrium, so test one of the outer charges.

Answer: (d) \( \dfrac35 \)

Explanation:
1. Initially \( F = \dfrac{kq_1q_2}{r^2} \). After contact each has \( \dfrac{q_1+q_2}{2} \), so \( F' = \dfrac{k(q_1+q_2)^2}{4r^2} \).
2. \( \dfrac{F'}{F} = \dfrac{(q_1+q_2)^2}{4q_1q_2} = \dfrac{16}{15} \). With \( x = q_1/q_2 \): \( 15(x+1)^2 = 64x \Rightarrow 15x^2-34x+15=0 \).
3. \( x = \dfrac{34\pm16}{30} = \dfrac53 \) or \( \dfrac35 \). So the ratio is 3 : 5 (or 5 : 3).

Teacher's Note:
1. The two roots are the same ratio read in opposite order, so only one of them appears among the options.

Answer: (c) \( \dfrac34F \)

Explanation:
1. Originally \( F = kq^2/r^2 \).
2. After transfer the charges are \( q/2 \) and \( q + q/2 = 3q/2 \).
3. \( F' = \dfrac{k(q/2)(3q/2)}{r^2} = \dfrac34F \).

Teacher's Note:
1. The total charge stays 2q, but the product of the charges decreases from q² to 3q²/4.

Answer: (c) \( r/\sqrt K \)

Explanation:
1. In the medium the force is \( F = \dfrac{q_1q_2}{4\pi\varepsilon_0K\,r'^2} \).
2. Setting this equal to the force in air: \( K r'^2 = r^2 \Rightarrow r' = \dfrac{r}{\sqrt K} \).

Teacher's Note:
1. The medium reduces the force K times, so the charges must be brought closer by √K to compensate.

Answer: (a) 30 cm from 9e

Explanation:
1. The third charge must lie between them, where the forces are opposite: \( \dfrac{9e}{x^2} = \dfrac{16e}{(70-x)^2} \).
2. \( \dfrac{x}{70-x} = \dfrac34 \Rightarrow x = 30 \) cm from 9e (which is 40 cm from 16e).

Teacher's Note:
1. For like charges the neutral point lies between them, closer to the smaller charge.

Answer: (c) decrease

Explanation:
1. On conducting spheres the like charges repel each other and redistribute towards the far sides.
2. The effective separation of the charge distributions is larger than 0.1 m, so the force decreases.

Teacher's Note:
1. Coulomb's law with centre-to-centre distance holds exactly only for uniformly distributed charge or for spheres that are far apart.

Answer: (b) 7:30

Explanation:
1. The +q at 12 pushes q0 towards 6, and the -q at 6 pulls it towards 6: both forces are along the 12-6 line towards 6.
2. The +q at 3 pushes q0 towards 9, and the -q at 9 pulls it towards 9.
3. The two equal components add to a force pointing midway between 6 and 9, which is the direction of the hour hand at 7:30.

Teacher's Note:
1. Resolve the four forces into two perpendicular directions. The resultant bisects the angle between them.

Answer: (b) \( 180^\circ,\ \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q^2}{(2L)^2} \)

Explanation:
1. With no gravity, the only forces are the repulsion and the tension, so the spheres move apart until the threads are in line: the angle is 180°.
2. The separation is then 2L and \( T = F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q^2}{(2L)^2} \).

Teacher's Note:
1. Without gravity there is nothing to balance a sideways component, so the threads must straighten out.

Answer: (c) \( \left(\dfrac{r}{\sqrt[3]{2}}\right) \)

Explanation:
1. For a small angle, \( \tan\theta = \dfrac{F}{mg} \Rightarrow \dfrac{r/2}{y} = \dfrac{kq^2}{r^2mg} \), so \( r^3 \propto y \).
2. With y halved: \( r'^3 = \dfrac{r^3}{2} \), so \( r' = \dfrac{r}{2^{1/3}} \).

Teacher's Note:
1. Write the equilibrium condition once, then compare the two cases using the proportionality r³ ∝ y.

Answer: (c) 3 N/C

Explanation:
1. Between oppositely charged parallel sheets, \( E = \dfrac{\sigma}{\varepsilon_0} \).
2. \( E = \dfrac{26.4\times10^{-12}}{8.85\times10^{-12}} \approx 3 \) N/C.

Teacher's Note:
1. For two sheets with equal and opposite charge the field is σ/ε0 between them and zero outside. For one sheet alone it would be σ/2ε0.

Answer: (d) \( \dfrac{Q}{2\pi^2\varepsilon_0R^2}(-\hat j) \)

Explanation:
1. The linear density is \( \lambda = \dfrac{Q}{\pi R} \). The components along x cancel by symmetry.
2. \( E_y = \int \dfrac{k\lambda R\,d\theta}{R^2}\sin\theta = \dfrac{k\lambda}{R}\cdot2 = \dfrac{2kQ}{\pi R^2} = \dfrac{Q}{2\pi^2\varepsilon_0R^2} \).
3. For a positive charge on the upper half, the field at O points away from the arc, along \( -\hat j \).

Teacher's Note:
1. Use symmetry to drop the component along the axis, then integrate the component perpendicular to it.

Answer: (b) \( \dfrac{\lambda_1r}{\lambda_1+\lambda_2} \)

Explanation:
1. The field of a line charge is \( E = \dfrac{\lambda}{2\pi\varepsilon_0x} \). Between the wires the fields are opposite.
2. Zero field: \( \dfrac{\lambda_1}{x} = \dfrac{\lambda_2}{r-x} \Rightarrow x = \dfrac{\lambda_1r}{\lambda_1+\lambda_2} \).

Teacher's Note:
1. The line-charge field falls as 1/distance, not 1/distance squared, so the null point is shifted accordingly.

Answer: (d) \( \dfrac{q}{\varepsilon_0\pi^2r^2} \) perpendicular to the line OP and directed downward

Explanation:
1. Each quarter-circle arc (charge q, radius r) subtends 90° at P. The field magnitude of such an arc is \( \dfrac{2kq\sqrt2}{\pi r^2} \), directed along the bisector of the arc.
2. The fields due to +q (away from the arc) and -q (towards the arc) have equal components along OP that cancel, and equal components perpendicular to OP, directed downward, that add.
3. \( E = 2\times\dfrac{2\sqrt2kq}{\pi r^2}\cos45^\circ = \dfrac{4kq}{\pi r^2} = \dfrac{q}{\pi^2\varepsilon_0r^2} \), perpendicular to OP, pointing downward.

Teacher's Note:
1. Split the semicircle into two quarter arcs, find the field of each, and add the components by symmetry.

Answer: (b) \( \dfrac{Q\,dl}{8\pi^2\varepsilon_0r^3} \)

Explanation:
1. The complete ring gives zero field at the centre, so the remaining wire gives a field equal and opposite to that of the removed piece.
2. The removed piece carries \( dq = \dfrac{Q}{2\pi r}dl \), so \( E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{r^2} = \dfrac{Q\,dl}{8\pi^2\varepsilon_0r^3} \).

Teacher's Note:
1. Superposition: (full ring) = (remaining wire) + (cut piece), and the full ring gives zero at the centre.

Answer: (d) \( 4\pi\varepsilon_0Q\times10^{22} \) volts/m

Explanation:
1. \( V = \dfrac{Q}{4\pi\varepsilon_0r} = Q\times10^{11} \Rightarrow r = \dfrac{1}{4\pi\varepsilon_0\times10^{11}} \).
2. \( E = \dfrac{V}{r} = Q\times10^{11}\times4\pi\varepsilon_0\times10^{11} = 4\pi\varepsilon_0Q\times10^{22} \) V/m.

Teacher's Note:
1. For a point charge, E = V/r. Find r from the given potential first.

Answer: (a) \( \hat i\,2xy+\hat j\,x^2-4\hat k \)

Explanation:
1. \( \vec E = -\nabla V = -\left(\hat i\dfrac{\partial V}{\partial x}+\hat j\dfrac{\partial V}{\partial y}+\hat k\dfrac{\partial V}{\partial z}\right) \).
2. \( \dfrac{\partial V}{\partial x} = -2xy,\ \dfrac{\partial V}{\partial y} = -x^2,\ \dfrac{\partial V}{\partial z} = 4 \).
3. \( \vec E = 2xy\,\hat i + x^2\hat j - 4\hat k \).

Teacher's Note:
1. The field is the negative gradient of the potential. Do not drop the minus sign.

Answer: (c) zero

Explanation:
1. A is at distance L from the two left charges (+q and -q) and at distance \( \sqrt{L^2+(2L)^2} = \sqrt5L \) from each right charge (-q and +q).
2. \( V = k\left[\dfrac qL-\dfrac qL-\dfrac{q}{\sqrt5L}+\dfrac{q}{\sqrt5L}\right] = 0 \).

Teacher's Note:
1. Potential is a scalar: equal and opposite charges at equal distances cancel pair by pair.

Answer: (d) 16 along negative X-axis

Explanation:
1. \( E_x = -\dfrac{dV}{dx} = -16x \).
2. At x = 1, \( E = -16 \) V/m, i.e. 16 V/m along the negative X-axis.

Teacher's Note:
1. A negative sign in the answer is the direction. Here the potential rises with x, so the field points the other way.

Answer: (a) \( Q = -q \)

Explanation:
1. All corners are at the same distance d from the centre.
2. \( V = \dfrac{k}{d}(-Q-q+2q+2Q) = \dfrac{k}{d}(Q+q) = 0 \Rightarrow Q = -q \).

Teacher's Note:
1. At the centre of a square every charge is at the same distance, so just add the charges algebraically.