Chapter 2 Electrostatics: Problem 2.11

Problem 2.11: Use Gauss’s law to find the electric field inside and outside a spherical
shell of radius R that carries a uniform surface charge density σ. Compare your
answer to Prob. 2.7.

Solution: 

 

 

 

 

 

 

 

 

 

 

Consider a gaussian spherical surface of radius $r$ where $r\ge R$.

According to Gauss Law, 

$$\oint \overrightarrow{E}\cdot \overrightarrow{da}=\frac{Q_{enc}}{{ϵ}_0}=\frac{\sigma \times 4\pi R^2}{{ϵ}_0}$$ 

Since the charge distribution is symmetrical, $\overrightarrow{E}$ and $\overrightarrow{da}$ point in the same direction and the direction is $\hat{r}$ 

$$E\times 4\pi r^2=\frac{\sigma \times 4\pi R^2}{{ϵ}_0}$$ 

$$\overrightarrow{E}=\frac{\sigma \times R^2}{{ϵ}_0\times r^2}\hat{r}$$ 

Now, For $r<R,Q_{enc}=0$
$$\therefore \oint \overrightarrow{E}\cdot \overrightarrow{da}=\frac{Q_{enc}}{{ϵ}_0}=0$$

Since, $\overrightarrow{da}\ne 0$ and both $\overrightarrow{E}$ and $\overrightarrow{da}$ points in the same direction.

$$\therefore \overrightarrow{E}=0$$

Hence ,

For $r<R,\overrightarrow{E}=0$ and

For $r\ge R$, $\overrightarrow{E}=\frac{\sigma \times R^2}{{ϵ}_0\times r^2}\hat{r}$  

 

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