Question

A parallel-plate capacitor is formed from two $6.0$-$cm$-diameter electrodes spaced $2.0mm$apart. The electric field strength inside the capacitor is $1.0\times {10}^{6}N/C$. What is the charge (in $nC$ ) on each electrode?

Short answer

One electrode has $25nC$charge and other one has $-25nC$charge.

Step-by-step solution

Introduction

Electric field $E$inside circuit is: $1\times {10}^{6}N/C$. The plate of a circuit has a diameter of$6cm$and they are $20cm$distance apart.

Formula used

Due to the area capacitor, there is an electric field. $A$and charge $Q$

$E=\frac{Q}{{ϵ}_{o}A}$...........$\left(1\right)$$\left(1\right)$

Explanation

Radius of disk is

$r=\frac{6cm}{2}$

$=3cm$

Now sum charge from equation $\left(1\right)$

$Q=E{ϵ}_{o}A$

$=\left(1\times {10}^6\mathrm{N}/\mathrm{C}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)\left(\pi (0.03\mathrm{m}{)}^2\right)$

$=25nC$

Find the charge on one electrode

The charge on one electrode is $25nC$, whereas the other is $-25nC$.