Question

What is the potential difference between the plates when the charge on the capacitor plates is 4.0 μC ?

Answer 1

Answer:

$1.34\times 10^4 V$

Explanation:

We are given that

The separation between parallel plate capacitor=d=0.81 mm=$0.81\times 10^{-3} m$

$1mm=10^{-3} m$

Area =A=$1.3\times 10^{-2} m^2$

Dielectric constant k=2.1

Charge on parallel plate capacitor=$Q=4.0\mu C=4\times 10^{-6} C$

$1\mu C=10^{-6} C$

We know that

$V=\frac{Qd}{k\epsilon_0A}$

Where $\epsilon_0=8.85\times 10^{-12}F/m$

Substitute the values

$V=\frac{4\times 10^{-6}\times 0.81\times 10^{-3}}{2.1\times 8.85\times 10^{-12}\times 1.3\times 10^{-2}} V$

$V=1.34\times 10^4 V$

Hence, the potential difference between the plates =$1.34\times 10^4 V$