Animated Solution for Physics - Electrostatics: A uniformly charged thin spherical shell of radius $R$ carries uniform surface charge density of $\sigma$ per unit area. It is made of two hemispherical shells, held together by pressing them with force $F$ (see figure). $F$ is proportional to

Imagine a thin spherical shell carrying a uniform positive charge. Because like charges repel, every small part of the shell experiences an outward push from the rest of the charge. To keep the two halves from flying apart, an external force $F$ is applied. Let's dive into the elegant physics that determines exactly how much force is needed!

Whenever we have a charged conductor, the charges reside on its surface and repel each other. This mutual repulsion creates an outward force per unit area, which we call electrostatic pressure ($P$).

Using Gauss's Law, we know that the electric field just outside the surface of a conductor is $E=\frac{\sigma }{{\epsilon }_0}$. The electrostatic pressure is given by the energy density of the electric field at the surface:

This pressure acts radially outward at every point on the spherical shell.

Let's focus on the left hemisphere. Consider a tiny area element $dS$ on its surface. The outward electrostatic force on this element is $d{F}_e=P\cdot dS$.

Because the hemisphere is curved, $d{F}_e$ points in different directions for different elements. If we resolve this force into horizontal ($x$) and vertical ($y$) components, something beautiful happens. Due to the perfect symmetry of the hemisphere, for every element in the upper half pushing upwards, there is a corresponding element in the lower half pushing downwards.

Since only the $x$-components survive, the net electrostatic force pushing the left hemisphere to the left is the sum of all the $x$-components of the force:

Because the pressure $P$ is uniform, we can pull it out of the integral:

Now, what is $\int (dS{)}_x$? It is the sum of the $x$-components of all the tiny area elements. Geometrically, this is exactly the projected area of the hemisphere onto the $y$-$z$ plane. Imagine shining a flashlight directly at the open face of the hemisphere; the shadow it casts is a flat circle of radius $R$.

Therefore, the projected area is simply $\pi R^2$.

Substituting the projected area back into our force equation, we get the total outward electrostatic force:

For the hemisphere to remain in equilibrium, the external force $F$ pushing it to the right must exactly balance this electrostatic force pushing it to the left.

From this, we can clearly see the proportionality:

This elegant use of projected area saves us from doing complex surface integrals and is a powerful tool in both electrostatics and fluid mechanics!