Question

When a potential difference of 154 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 29.0 nC/cm2. What is the spacing between the plates?

Answer 1

Answer:

$4.7\mu m$

Explanation:

We are given that

Potential difference=V=154 V

Surface charge density=$\sigma=29nC/m^2=29\times 10^{-5} C/m^2$

Using $1 nC/cm^2=10^{-5} C/m^2$

We know that

$C=\frac{\epsilon_0A}{d}=\frac{Q}{V}=\frac{\sigma A}{V}$

$d=\frac{\epsilon_0V}{\sigma}$

Where

$\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Using the formula

$d=\frac{8.85\times 10^{-12}\times 154}{29\times 10^{-5}}$

$d=4.7\times 10^{-6} m=4.7\mu m$

Using $1\mu m=10^{-6} m$

Hence, the spacing between the plates=$4.7\mu m$