Electrostatics — JEE Main practice

1. The work done in moving a charge equals the charge times the potential difference between the final and initial points, so first find the distances of points A and B from the origin. Given point A(4,4,2),

$$r_A = \sqrt{4^2+4^2+2^2} = \sqrt{36} = 6\ \text{m}$$

Given point B(2,2,1),

$$r_B = \sqrt{2^2+2^2+1^2} = \sqrt{9} = 3\ \text{m}$$

2. Compute the potential at each point due to the source charge at the origin. Given $Q=10^{-8}$ C,

$$V_A = \frac{kQ}{r_A} = \frac{(9\times10^9)(10^{-8})}{6} = 15\ \text{V}$$ $$V_B = \frac{kQ}{r_B} = \frac{(9\times10^9)(10^{-8})}{3} = 30\ \text{V}$$

3. Use $W=q(V_B-V_A)$ to find the work done in moving the test charge from A to B. Given $q=2\times10^{-6}$ C,

$$W = (2\times10^{-6})(30-15) = 30\times10^{-6}\ \text{J}$$

Hence, the work done is $30\times10^{-6}$ J, so the answer is option A.

1. Since both electrostatic and gravitational forces act between the spheres, use total energy conservation (kinetic + electric potential + gravitational potential energy) between the initial and final states. Given each sphere moves with speed $u$, the initial kinetic energy for both spheres is

$$K_i = mu^2$$

2. Compute the initial potential energies (electric and gravitational) with centres separated by $4R$. Given

$$U_{ei} = \frac{kQ^2}{4R}, \qquad U_{gi} = -\frac{Gm^2}{4R}$$

3. At the moment the spheres just touch, their centres are separated by $2R$ (sum of radii), and for minimum speed we assume they momentarily come to rest there. Therefore, $K_f=0$, and

$$U_{ef} = \frac{kQ^2}{2R}, \qquad U_{gf} = -\frac{Gm^2}{2R}$$

4. Apply energy conservation between initial and final states. Given

$$K_i+U_{ei}+U_{gi} = K_f+U_{ef}+U_{gf}$$ $$mu^2+\frac{kQ^2}{4R}-\frac{Gm^2}{4R} = \frac{kQ^2}{2R}-\frac{Gm^2}{2R}$$

5. Rearrange to isolate $u^2$. Therefore,

$$mu^2 = \frac{kQ^2}{4R}-\frac{Gm^2}{4R} = \frac{kQ^2}{4R}\left(1-\frac{Gm^2}{kQ^2}\right)$$

6. Solve for $u$. Hence,

$$u = \sqrt{\frac{kQ^2}{4mR}\left(1-\frac{Gm^2}{kQ^2}\right)}$$

Hence, the minimum speed is as given, so the answer is option D.

1. Each pair of charges directly opposite each other on the circle (separated by 180°) produce equal and opposite fields at the centre, so identify which pairs cancel out first. The charges at 90° and 270° cancel each other, and similarly the charges at 30° and 210° cancel each other, leaving only the pair at 150° and 330° (or equivalently, an uncancelled pair) contributing to the net field.

2. For each remaining charge, the field magnitude at the centre from a single charge $Q$ at distance $R$ is

$$E_0 = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}$$

3. Since the two remaining contributions add constructively along the same direction, the net magnitude is twice $E_0$. Therefore,

$$|\vec{E}_{net}| = 2E_0$$

4. Determine the direction of this net vector using the given geometry (found to be along $150°$ from the positive x-axis) and resolve into components. Hence,

$$\vec{E}_{net} = 2E_0(\cos150°\hat{i}+\sin150°\hat{j}) = -E_0(\sqrt3\hat{i}-\hat{j})$$

5. Substitute back $E_0=\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}$. Hence,

$$\vec{E}_{net} = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(\sqrt3\hat{i}-\hat{j})$$

Hence, the answer is option B.