# Breakdown Voltage of a Sphere

What is the maximum charge that can be given to a sphere of diameter (10cm) if the breakdown voltage of air is (2\times10^4 V/cm)?

Discussion:

Dielectric breakdown is when current flows through an electrical insulator when the voltage applied across it exceeds the breakdown voltage. This results in the insulator becoming electrically conductive.

The breakdown voltage of air is (2\times10^4 V/cm)

For a sphere, $$V= \frac{q}{4\pi \epsilon_0 r}$$
We can find charge $$q= 4\pi \epsilon_0 r V$$ $$= \frac{(5)(2\times10^4)}{9\times10^9}$$ $$=1.11\times 10^{-5}$$

# Capacitance of a Spherical Capacitor

Try this problem, useful for Physics Olympiad, based on Capacitance of a Spherical Capacitor.

The Problem:

A spherical capacitor has inner radius (a) and outer radius (b). It is filled with an inhomogeneous dielectric with permittivity $$\epsilon=\epsilon_0K/r^2$$ for (a<r<b). The outer sphere is grounded and a charge is placed on the inner sphere. Find the capacitance of the system.

Solution:
The electric field at any inside point is $$\vec{E}=\frac{Q}{4\pi\epsilon r^2}\hat{r} =\frac{Q}{4\pi\epsilon_0K}\hat{r}$$ where (Q) is the charge on the inner sphere.
Now, the potential difference between the spheres is $$V=-\int_{b}^{a} \vec{E}.\vec{dr}$$$$=\int_{a}^{b}\frac{Q}{4\pi\epsilon_0K}dr$$$$=\frac{Q}{4\pi\epsilon_0K}(b-a)$$
Capacitance $$C=\frac{Q}{V}=\frac{4\pi\epsilon_0K}{b-a}$$