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

`4.7\mu m`

Step-by-step explanation:

We are given that

Potential difference=V=154 V

Surface charge density=
`\sigma=29nC/m^2=29* 10^(-5) C/m^2`

Using
`1 nC/cm^2=10^(-5) C/m^2`

We know that

`C=(\epsilon_0A)/(d)=(Q)/(V)=(\sigma A)/(V)`

`d=(\epsilon_0V)/(\sigma)`

Where

`\epsilon_0=8.85* 10^(-12)C^2/Nm^2`

Using the formula

`d=(8.85* 10^(-12)* 154)/(29* 10^(-5))`

`d=4.7* 10^(-6) m=4.7\mu m`

Using
`1\mu m=10^(-6) m`

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