Electric Potential and Field due to Charged Spherical Shell with a Hole

To determine the correct statement, we use the principle of superposition.

1. Initial Setup and Superposition Principle

Let the original complete insulating spherical shell have radius $R$ and uniform surface charge density $\sigma$. The total charge on the complete shell is: $Q = \sigma (4\pi R^2)$

The potential at the surface of this full shell is given as: $V_0 = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$

When a tiny hole of area $\Delta A = \alpha (4\pi R^2)$ (where $\alpha \ll 1$) is removed, the charge removed is: $\Delta q = \sigma \Delta A = \sigma \alpha (4\pi R^2) = \alpha Q$

By the principle of superposition, the electric field and potential of the shell with the hole can be modeled as the combination of:

1. A complete, uniformly charged spherical shell with total charge $Q$.
2. A point-like patch with charge $-\Delta q = -\alpha Q$ located at the position of the hole, $\vec{r}_h = R\hat{k}$ (choosing the $z$-axis along the line from the center through the hole).

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2. Evaluation of Options

Analysis of Option A:

The potential at the center $C$ ($\vec{r} = 0$) is: $V_C = V_{\text{full shell}} + V_{\text{patch}}$ $V_C = V_0 + \frac{1}{4\pi\varepsilon_0} \frac{-\alpha Q}{R} = V_0 - \alpha V_0 = V_0 (1 - \alpha)$

The reduction in potential at the center is $\alpha V_0$. Therefore, Option A (which claims a reduction of $2\alpha V_0$) is incorrect.

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Analysis of Option B:

The electric field at the center due to the full shell is zero ($\vec{E}_{\text{full}} = 0$). The net electric field at the center is entirely due to the removed patch: $\vec{E}_C = \vec{E}_{\text{full}} + \vec{E}_{\text{patch}} = 0 + \frac{1}{4\pi\varepsilon_0} \frac{-\alpha Q}{R^2} (-\hat{k}) = \frac{\alpha Q}{4\pi\varepsilon_0 R^2} \hat{k} = \frac{\alpha V_0}{R} \hat{k}$

The magnitude of the electric field at the center increases from $0$ to $\frac{\alpha V_0}{R}$. Therefore, Option B is incorrect.

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Analysis of Option C:

Let point $P$ be located at a distance $r = \frac{1}{2}R$ from the center along the direction towards the hole, so its position vector is $\vec{r}_P = \frac{1}{2}R \hat{k}$.

• Since $P$ lies inside the complete shell, the potential due to the full shell at $P$ is: $V_{\text{full at } P} = V_0$

• The distance from point $P$ to the hole at $R\hat{k}$ is: $d_P = R - \frac{1}{2}R = \frac{1}{2}R$

• The potential at point $P$ due to the patch is: $V_{\text{patch at } P} = \frac{1}{4\pi\varepsilon_0} \frac{-\alpha Q}{d_P} = \frac{1}{4\pi\varepsilon_0} \frac{-\alpha Q}{R/2} = -2\alpha \left(\frac{1}{4\pi\varepsilon_0} \frac{Q}{R}\right) = -2\alpha V_0$

Thus, the total potential at point $P$ is: $V_P = V_{\text{full at } P} + V_{\text{patch at } P} = V_0 - 2\alpha V_0 = V_0 (1 - 2\alpha)$

Now, computing the ratio of the potential at the center $V_C$ to the potential at $P$: $\frac{V_C}{V_P} = \frac{V_0(1 - \alpha)}{V_0(1 - 2\alpha)} = \frac{1 - \alpha}{1 - 2\alpha}$

Therefore, Option C is correct.

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Analysis of Option D:

For a point at distance $2R$ from the center along the axis towards the hole ($\vec{r} = 2R\hat{k}$):

• Initial field magnitude: $E_{\text{initial}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{(2R)^2} = \frac{V_0}{4R}$
• Distance to the hole: $d = 2R - R = R$
• Field due to patch: $\vec{E}_{\text{patch}} = -\frac{1}{4\pi\varepsilon_0}\frac{\alpha Q}{R^2}\hat{k} = -\frac{\alpha V_0}{R}\hat{k}$

The field reduction magnitude is $\frac{\alpha V_0}{R}$, not $\frac{\alpha V_0}{2R}$. Therefore, Option D is incorrect.

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Conclusion

The correct statement is C.