Question

An electric field of 8.5X10^5 V/m is desired between two parallel plates each of areas 2500 cm^2 and separated by 0.10 mm of air. What charge must be on each plate?

Answer 1

Answer:

The charge must be on each plate is $1.881\times10^{-6}\ C$.

Explanation:

Given that,

Electric field $E= 8.5\times10^{5}\ V/m$

Area = 2500 cm²

Distance = 0.10 mm

We need to calculate the potential difference

Using formula of potential difference

$\Delta V=E\times d$

$\Delta V=8.5\times10^{5}\times0.10\times10^{-3}$

$\Delta V=85\ V$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}\times A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times2500\times10^{-4}}{0.10\times10^{-3}}$

$C=2.2125\times10^{-8}\ F$

We need to calculate the charge

Using formula of charge

$Q=C\times\Delta V$

$Q=2.2125\times10^{-8}\times85$

$Q=1.881\times10^{-6}\ C$

Hence, The charge must be on each plate is $1.881\times10^{-6}\ C$.