Question

Magnitude of surface charge density on each plate of a parallel plate capacitor is $150{\epsilon }_0\mu {\text{C/m}}^2$. What is the magnitude of the electric field in the region between the plates?
(Take ${\epsilon }_o=8.85\times {10}^{-12}{\text{C}}^2{\text{/N m}}^2$)

Answer 1

Given:
surface charge density, $\sigma =150{\epsilon }_0\mu {\text{C/m}}^2$

The electric field due to each plate has the same magnitude and direction.

$\left|{\overrightarrow{E}}_{+}\right|=\left|{\overrightarrow{E}}_{-}\right|=\frac{\sigma }{2{ϵ}_0}\text{[towards right]}$

$\therefore \left|{\overrightarrow{E}}_{net}\right|=\left|{\overrightarrow{E}}_{+}\right|+\left|{\overrightarrow{E}}_{-}\right|$

$\Rightarrow \left|{\overrightarrow{E}}_{net}\right|=\frac{\sigma }{2{ϵ}_0}+\frac{\sigma }{2{ϵ}_0}=\frac{\sigma }{{ϵ}_0}$

$\Rightarrow \left|{\overrightarrow{E}}_{net}\right|=\frac{150\times 8.85\times {10}^{-6}}{8.85\times {10}^{-12}}=150\text{N/C}$

Key Concept : Magnitude of electric field due to each plate is $\frac{\sigma }{2{ϵ}_0}$ and it's directed along the same direction, which is from positive plate towards negative plate.