Question

Consider a uniformly charged spherical shell. Two cones having same semi vertical angle, and their common apex at $P$, intercept the shell. The intercepts have area $\Delta S1$ and $\Delta S2$. For a cone of very small angle, $\Delta S1$ and $\Delta S2$ will be very small and charge on them can be regarded as point charge for the purpose of writing electric field at point $P$. Prove that the charge on $\Delta S1$ and $\Delta S2$ produce equal and opposite field at $P$. Hence, argue that field at all points inside the uniformly charged spherical shell is zero.

Answer 1

The charge on $\Delta S1$ and $\Delta S2$ produce equal and opposite field at $P$. Hence, the field at all points inside the uniformly charged spherical shell is zero.

Answer 2

Let $P$ be a point inside the uniformly charged spherical shell. Consider a small cone with apex at $P$ and a small solid angle $\Delta \Omega$. This cone intercepts two small areas $\Delta S_1$ and $\Delta S_2$ on the spherical shell. Let the distances from $P$ to these areas be $r_1$ and $r_2$, respectively. Since the cone is very small, $\Delta S_1$ and $\Delta S_2$ can be considered as approximately planar.

The solid angle $\Delta \Omega$ subtended by an area element $\Delta S$ at a point $P$ at a distance $r$ is given by $\Delta \Omega = \frac{\Delta S \cos \alpha}{r^2}$, where $\alpha$ is the angle between the normal to the area element and the line connecting $P$ to the area element.

For the area $\Delta S_1$, let the angle between the line from $P$ to $\Delta S_1$ and the outward normal to the shell at $\Delta S_1$ be $\alpha_1$. Then $\Delta \Omega = \frac{\Delta S_1 \cos \alpha_1}{r_1^2}$.

For the area $\Delta S_2$, let the angle between the line from $P$ to $\Delta S_2$ and the outward normal to the shell at $\Delta S_2$ be $\alpha_2$. Then $\Delta \Omega = \frac{\Delta S_2 \cos \alpha_2}{r_2^2}$.

Since the solid angle is the same for both intercepts, we have $\frac{\Delta S_1 \cos \alpha_1}{r_1^2} = \frac{\Delta S_2 \cos \alpha_2}{r_2^2}$.

Consider the line passing through $P$ and the centers of $\Delta S_1$ and $\Delta S_2$. Let this line intersect the spherical shell at points $A$ and $B$, where $A$ is associated with $\Delta S_1$ and $B$ is associated with $\Delta S_2$. Let $O$ be the center of the spherical shell. The outward normal at any point on the shell is along the line connecting that point to $O$. Let the line segment $APB$ make an angle $\theta$ with the line $PO$.

Consider the angles $\alpha_1$ and $\alpha_2$. It can be shown that $\cos \alpha_1 = \cos \alpha_2$.

We have $\frac{\Delta S_1}{r_1^2 \cos \alpha_1} = \frac{\Delta S_2}{r_2^2 \cos \alpha_2}$.

Let's use the fact that $\Delta \Omega = \frac{\Delta S \cos \alpha}{r^2}$.

We need to show $\frac{\Delta S_1}{r_1^2} = \frac{\Delta S_2}{r_2^2}$. This is equivalent to showing $\frac{\cos \alpha_1}{\Delta \Omega} = \frac{\cos \alpha_2}{\Delta \Omega}$, which means $\cos \alpha_1 = \cos \alpha_2$.

From $\frac{\Delta S_1 \cos \alpha_1}{r_1^2} = \frac{\Delta S_2 \cos \alpha_2}{r_2^2}$ and $\cos \alpha_1 = \cos \alpha_2 \ne 0$, we get $\frac{\Delta S_1}{r_1^2} = \frac{\Delta S_2}{r_2^2}$.

The charge on $\Delta S_1$ is $\Delta q_1 = \sigma \Delta S_1$. The electric field at $P$ due to $\Delta q_1$ is $\Delta \vec{E}_1 = \frac{1}{4\pi\epsilon_0} \frac{\Delta q_1}{r_1^2} \hat{r}_1 = \frac{1}{4\pi\epsilon_0} \frac{\sigma \Delta S_1}{r_1^2} \hat{r}_1$.

The charge on $\Delta S_2$ is $\Delta q_2 = \sigma \Delta S_2$. The electric field at $P$ due to $\Delta q_2$ is $\Delta \vec{E}_2 = \frac{1}{4\pi\epsilon_0} \frac{\Delta q_2}{r_2^2} \hat{r}_2 = \frac{1}{4\pi\epsilon_0} \frac{\sigma \Delta S_2}{r_2^2} \hat{r}_2$.

Since the points are on a line through $P$, the unit vectors are opposite in direction, i.e., $\hat{r}_2 = -\hat{r}_1$.

So, $\Delta \vec{E}_1 = \frac{\sigma}{4\pi\epsilon_0} \frac{\Delta S_1}{r_1^2} \hat{r}_1$ and $\Delta \vec{E}_2 = \frac{\sigma}{4\pi\epsilon_0} \frac{\Delta S_2}{r_2^2} (-\hat{r}_1)$.

Since $\frac{\Delta S_1}{r_1^2} = \frac{\Delta S_2}{r_2^2}$, we have $|\Delta \vec{E}_1| = \frac{\sigma}{4\pi\epsilon_0} \frac{\Delta S_1}{r_1^2}$ and $|\Delta \vec{E}_2| = \frac{\sigma}{4\pi\epsilon_0} \frac{\Delta S_2}{r_2^2} = \frac{\sigma}{4\pi\epsilon_0} \frac{\Delta S_1}{r_1^2} = |\Delta \vec{E}_1|$.

Also, $\Delta \vec{E}_2 = -\frac{\sigma}{4\pi\epsilon_0} \frac{\Delta S_1}{r_1^2} \hat{r}_1 = -\Delta \vec{E}_1$.

Thus, the charge on $\Delta S_1$ and $\Delta S_2$ produce equal and opposite electric fields at $P$.

To argue that the field at all points inside the uniformly charged spherical shell is zero, we can consider any point $P$ inside the shell. We can divide the entire spherical shell into pairs of areas intercepted by small cones with apex at $P$ and the same solid angle. For each pair of areas $\Delta S_1$ and $\Delta S_2$ intercepted by a cone, the electric fields produced at $P$ are equal in magnitude and opposite in direction, so they cancel each other out. By summing over all such pairs of areas that make up the entire spherical shell, the net electric field at $P$ is the vector sum of all these canceling pairs, which is zero. Since this holds for any point $P$ inside the spherical shell, the electric field at all points inside the uniformly charged spherical shell is zero.