A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential at its surface is V0. A hole with a small area \alpha 4 \pi R2( \alpha << 1) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

The ratio of the potential at the center of the shell of that of the point at 1 \over 2 R from center towards the hole will be 1 - \alpha \over 1 - 2\alpha .

The potential at the center of the shell is reduced by 2 \alpha V0.

The magnitude of electric field at the center of the shell is reduced by \alpha V_0 \over 2R .

The magnitude of electric field at a point, located on a line passing through the hole and shell's center, on a distance 2R from the center of the spherical shell will be reduced by \alpha V_0 \over 2R .

Electrostatics PYQ — JEE Advanced 2019 · Shift 1 · Q39 with Solution

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Correct answer: A

Initial charge on the complete shell

For a uniformly charged thin spherical shell of radius $R$ , $V_0 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$ So, $Q = 4\pi\varepsilon_0 R V_0$

The surface charge density on the original shell is $\sigma = \frac{Q}{4\pi R^2}$

Charge removed due to the small hole

The hole area is $A_h = \alpha \cdot 4\pi R^2, \qquad \alpha \ll 1$

Hence removed charge is $q = \sigma A_h = \frac{Q}{4\pi R^2}(\alpha 4\pi R^2) = \alpha Q$

So the new configuration can be treated as:

full uniformly charged shell of charge $Q$
plus a small patch of charge $-q=-\alpha Q$ at the hole location

Since $\alpha \ll 1$ , the missing patch may be approximated as a point charge $-\alpha Q$ at the surface point where the hole is made.

Potential at the center

Potential at center due to full shell is $V_0$ .

Distance of the center from the hole-point is $R$ , so potential due to missing charge is $V_{\text{missing at center}}=\frac{1}{4\pi\varepsilon_0}\frac{\alpha Q}{R}=\alpha V_0$

Since this charge is removed, the new potential at center is $V_c = V_0-\alpha V_0 = V_0(1-\alpha)$

So statement B says reduction is $2\alpha V_0$ , which is false.

Potential at the point $P$ located at $R/2$ from center towards the hole

For a complete shell, potential anywhere inside is constant, so contribution from full shell is again $V_0$

Now consider the removed patch approximated as a point charge $\alpha Q$ at the surface point of the hole.

Distance from this hole-point to $P$ : $R-\frac{R}{2}=\frac{R}{2}$

Thus potential contribution of removed charge at $P$ would have been

= \frac{2\alpha Q}{4\pi\varepsilon_0 R} = 2\alpha V_0$$ Hence after removing it, $$V_P = V_0 - 2\alpha V_0 = V_0(1-2\alpha)$$ Therefore, $$\frac{V_c}{V_P} = \frac{V_0(1-\alpha)}{V_0(1-2\alpha)} = \frac{1-\alpha}{1-2\alpha}$$ This matches **Option A**. --- 5. **Check electric field at the center** For a complete shell, field at center is zero. After making the hole, field at center is due to the missing patch, approximated as a point charge $\alpha Q$ at distance $R$: $$E_c = \frac{1}{4\pi\varepsilon_0}\frac{\alpha Q}{R^2}$$ Using $Q=4\pi\varepsilon_0 R V_0$, $$E_c = \frac{\alpha (4\pi\varepsilon_0 R V_0)}{4\pi\varepsilon_0 R^2} = \frac{\alpha V_0}{R}$$ So field changes from $0$ to $\frac{\alpha V_0}{R}$, not reduced by $\frac{\alpha V_0}{2R}$. Hence C is false. --- 6. **Check electric field at a point on the axis through hole and center, at distance $2R$ from center** Take the external point on the hole side. For complete shell, $$E_{\text{full}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{(2R)^2} = \frac{Q}{16\pi\varepsilon_0 R^2} = \frac{V_0}{4R}$$ The removed patch is approximately a point charge $\alpha Q$ at radius $R$, so its distance from this external point is $$2R-R=R$$ Hence reduction in field magnitude is $$\Delta E = \frac{1}{4\pi\varepsilon_0}\frac{\alpha Q}{R^2} = \frac{\alpha V_0}{R}$$ This is not $\frac{\alpha V_0}{2R}$, so D is false. --- 7. **Conclusion** Only option **A** is correct. $$\boxed{\text{A}}$$

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Correct answer: A

Initial charge on the complete shell

For a uniformly charged thin spherical shell of radius $R$ , $V_0 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$ So, $Q = 4\pi\varepsilon_0 R V_0$

The surface charge density on the original shell is $\sigma = \frac{Q}{4\pi R^2}$

Charge removed due to the small hole

The hole area is $A_h = \alpha \cdot 4\pi R^2, \qquad \alpha \ll 1$

Hence removed charge is $q = \sigma A_h = \frac{Q}{4\pi R^2}(\alpha 4\pi R^2) = \alpha Q$

So the new configuration can be treated as:

full uniformly charged shell of charge $Q$
plus a small patch of charge $-q=-\alpha Q$ at the hole location

Since $\alpha \ll 1$ , the missing patch may be approximated as a point charge $-\alpha Q$ at the surface point where the hole is made.

Potential at the center

Potential at center due to full shell is $V_0$ .

Distance of the center from the hole-point is $R$ , so potential due to missing charge is $V_{\text{missing at center}}=\frac{1}{4\pi\varepsilon_0}\frac{\alpha Q}{R}=\alpha V_0$

Since this charge is removed, the new potential at center is $V_c = V_0-\alpha V_0 = V_0(1-\alpha)$

So statement B says reduction is $2\alpha V_0$ , which is false.

Potential at the point $P$ located at $R/2$ from center towards the hole

For a complete shell, potential anywhere inside is constant, so contribution from full shell is again $V_0$

Now consider the removed patch approximated as a point charge $\alpha Q$ at the surface point of the hole.

Distance from this hole-point to $P$ : $R-\frac{R}{2}=\frac{R}{2}$

Thus potential contribution of removed charge at $P$ would have been

= \frac{2\alpha Q}{4\pi\varepsilon_0 R} = 2\alpha V_0$$ Hence after removing it, $$V_P = V_0 - 2\alpha V_0 = V_0(1-2\alpha)$$ Therefore, $$\frac{V_c}{V_P} = \frac{V_0(1-\alpha)}{V_0(1-2\alpha)} = \frac{1-\alpha}{1-2\alpha}$$ This matches **Option A**. --- 5. **Check electric field at the center** For a complete shell, field at center is zero. After making the hole, field at center is due to the missing patch, approximated as a point charge $\alpha Q$ at distance $R$: $$E_c = \frac{1}{4\pi\varepsilon_0}\frac{\alpha Q}{R^2}$$ Using $Q=4\pi\varepsilon_0 R V_0$, $$E_c = \frac{\alpha (4\pi\varepsilon_0 R V_0)}{4\pi\varepsilon_0 R^2} = \frac{\alpha V_0}{R}$$ So field changes from $0$ to $\frac{\alpha V_0}{R}$, not reduced by $\frac{\alpha V_0}{2R}$. Hence C is false. --- 6. **Check electric field at a point on the axis through hole and center, at distance $2R$ from center** Take the external point on the hole side. For complete shell, $$E_{\text{full}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{(2R)^2} = \frac{Q}{16\pi\varepsilon_0 R^2} = \frac{V_0}{4R}$$ The removed patch is approximately a point charge $\alpha Q$ at radius $R$, so its distance from this external point is $$2R-R=R$$ Hence reduction in field magnitude is $$\Delta E = \frac{1}{4\pi\varepsilon_0}\frac{\alpha Q}{R^2} = \frac{\alpha V_0}{R}$$ This is not $\frac{\alpha V_0}{2R}$, so D is false. --- 7. **Conclusion** Only option **A** is correct. $$\boxed{\text{A}}$$

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