A positive point charge of $$10^{-8}$$ C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved a distance of 10 cm further away from the center of the sphere along the radial direction. Taking $$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$$ Nm$$^2$$/C$$^2$$ (where $$\epsilon_0$$ is the permittivity of free space), which of the following statements is/are correct:

Given: point charge $$q = +10^{-8}\,\text{C}$$ at a distance $$d = 20\,\text{cm} = 0.20\,\text{m}$$ from the centre $$O$$ of a conducting sphere of radius $$R = 10\,\text{cm} = 0.10\,\text{m}$$.
Throughout we use $$k = \dfrac{1}{4\pi\varepsilon_0}=9\times10^9\;\text{N m}^2\!\text{/C}^2$$.

Step 1 : Image-charge data for a charge outside a sphere
For a point charge $$q$$ kept at distance $$d\,(d\gt R)$$ from the centre of a conducting sphere of radius $$R$$:

• An image charge $$q' = -\dfrac{R}{d}\,q$$ is located on the same line at a distance $$b = \dfrac{R^{2}}{d}$$ from the centre.
• The combination $$(q,q')$$ alone makes the potential of the sphere zero (grounded condition).
• If the sphere is isolated (not grounded) and must keep a specified net charge $$Q_{\text{sphere}}$$, an additional charge $$q_c$$ is placed at the centre. $$q_c$$ is fixed by the condition $$q_c + q' = Q_{\text{sphere}}$$. The potential of the sphere then becomes a constant $$V_{\text{sphere}} = k\dfrac{q_c}{R}$$.

Part (A) : Potential before grounding
Initially the sphere is neutral $$(Q_{\text{sphere}}=0)$$ and isolated.

Image charge: $$q' = -\dfrac{R}{d}\,q = -\dfrac{0.10}{0.20}\,10^{-8} = -0.5\times10^{-8} = -5\times10^{-9}\,\text{C}$$.
To keep the sphere neutral, add a central charge $$q_c = -q' = +5\times10^{-9}\,\text{C}$$.

The potential of the sphere is therefore $$V_{\text{before}} = k\dfrac{q_c}{R} = 9\times10^9 \times\dfrac{5\times10^{-9}}{0.10} = 9\times10^9 \times 5\times10^{-8} = 45\times10^{1} = 450\;\text{V}.$$

Hence Option A is correct.

Part (B) : Charge that flows when the sphere is grounded
On connecting the sphere to earth, its potential must become zero. That is achieved by removing the central charge $$q_c$$ (ground provides/absorbs charge) so that only $$q'$$ remains on the sphere.

Charge leaving the sphere to the ground $$\;\Delta Q = q_c = +5\times10^{-9}\,\text{C}.$$

Thus $$5\times10^{-9}\,\text{C}$$ of positive charge flows from the sphere to the earth. Option B is correct.

Part (C) : Charge on the isolated sphere after the grounding wire is removed
With the earth connection opened, the induced charge that remains on the sphere equals $$q' = -5\times10^{-9}\,\text{C}$$.

Option C is correct.

Part (D) : Potential after the external charge is shifted to 30 cm
The grounding has already been removed, so the total charge on the sphere stays fixed at $$Q_{\text{sphere}} = -5\times10^{-9}\,\text{C}.$$ Now the point charge is shifted to $$d_2 = 30\,\text{cm} = 0.30\,\text{m}$$.

New image charge for the new position:
$$q'_2 = -\dfrac{R}{d_2}\,q = -\dfrac{0.10}{0.30}\,10^{-8} = -\dfrac{1}{3}\times10^{-8} = -3.33\times10^{-9}\,\text{C}.$$

To keep the sphere’s total charge fixed, the central charge becomes $$q_{c2} = Q_{\text{sphere}} - q'_2 = \bigl(-5\times10^{-9}\bigr) - \bigl(-3.33\times10^{-9}\bigr) = -1.67\times10^{-9}\,\text{C}.$$

The potential of the sphere is therefore $$V_{\text{final}} = k\dfrac{q_{c2}}{R} = 9\times10^9 \times\dfrac{-1.67\times10^{-9}}{0.10} \approx -150\;\text{V}.$$

This is neither 300 V nor positive. Hence Option D is incorrect.

Result
Correct statements: Option A, Option B, Option C.