Electrostatic Potential and Induced Charge on Grounded Conducting Sphere

To determine which statements are correct, we analyze the electrostatic properties of the system at each step:

1. Potential of the sphere before grounding (Option A)

Before grounding, the conducting sphere is neutral, meaning its net charge is $Q_1 = 0$. Since the sphere is a conductor, it is an equipotential body, and its potential is equal to the electrostatic potential at its center $C$.

The total potential at the center $C$ is the sum of the potential due to the point charge $q = 10^{-8}\text{ C}$ located at distance $d_1 = 20\text{ cm} = 0.2\text{ m}$ and the potential due to the induced surface charges $q_{\text{ind}}$: $V_{\text{sphere}} = V_C = \frac{1}{4\pi\epsilon_0} \frac{q}{d_1} + V_{\text{induced charges at } C}$

Since all induced surface charges lie at a distance equal to the radius $R = 10\text{ cm} = 0.1\text{ m}$ from the center, their net contribution to the potential at $C$ is: $V_{\text{induced charges at } C} = \frac{1}{4\pi\epsilon_0} \frac{\int \text{d}q}{R} = \frac{1}{4\pi\epsilon_0} \frac{Q_{\text{net, induced}}}{R} = 0 \quad (\text{since } Q_{\text{net, induced}} = 0)$

Thus: $V_{\text{sphere}} = \frac{1}{4\pi\epsilon_0} \frac{q}{d_1} = \left(9 \times 10^9\text{ N m}^2/\text{C}^2\right) \times \frac{10^{-8}\text{ C}}{0.2\text{ m}} = 450\text{ V}$

Statement (A) is correct.

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2. Charge on the sphere after grounding and charge flow (Options B and C)

When the sphere is grounded, its electrostatic potential becomes zero ($V_{\text{sphere}}' = 0$).

Let $q'$ be the net charge on the sphere after grounding. The potential at the center of the sphere is now: $V_C' = \frac{1}{4\pi\epsilon_0} \frac{q}{d_1} + \frac{1}{4\pi\epsilon_0} \frac{q'}{R} = 0$

Substitute the given values: $\frac{q}{d_1} + \frac{q'}{R} = 0 \implies q' = -q \left(\frac{R}{d_1}\right)$ $q' = -\left(10^{-8}\text{ C}\right) \times \left(\frac{0.1\text{ m}}{0.2\text{ m}}\right) = -5 \times 10^{-9}\text{ C}$

• Charge flowing to the ground: The change in charge on the sphere is: $\Delta q = q' - 0 = -5 \times 10^{-9}\text{ C}$ Therefore, a positive charge of $5 \times 10^{-9}\text{ C}$ flows from the sphere to the ground (or $5 \times 10^{-9}\text{ C}$ of negative charge flows from the ground to the sphere).

  Statement (B) is correct.

• Charge remaining on the sphere: After the grounding wire is disconnected, the charge remaining on the isolated sphere is $q' = -5 \times 10^{-9}\text{ C}$.

  Statement (C) is correct.

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3. Final electrostatic potential of the sphere (Option D)

After removing the ground connection, the point charge $q = 10^{-8}\text{ C}$ is moved by an additional distance of $10\text{ cm}$ radially away from the center. The new distance from the center is: $d_2 = 20\text{ cm} + 10\text{ cm} = 30\text{ cm} = 0.3\text{ m}$

The charge on the sphere remains constant at $q' = -5 \times 10^{-9}\text{ C}$.

The new electrostatic potential of the sphere is calculated at its center: $V_{\text{final}} = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{d_2} + \frac{q'}{R} \right)$ $V_{\text{final}} = \left(9 \times 10^9\right) \left( \frac{10^{-8}}{0.3} + \frac{-5 \times 10^{-9}}{0.1} \right)$ $V_{\text{final}} = 90 \left( \frac{10}{3} - 5 \right) = 90 \left( -\frac{5}{3} \right) = -150\text{ V}$

Thus, the final electrostatic potential of the sphere is $-150\text{ V}$, not $300\text{ V}$.

Statement (D) is incorrect.

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Conclusion

The correct choices are A, B, and C.