Question

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

Answer 1

`1.34* 10^4 V`

Step-by-step explanation:

We are given that

The separation between parallel plate capacitor=d=0.81 mm=
`0.81* 10^(-3) m`

`1mm=10^(-3) m`

Area =A=
`1.3* 10^(-2) m^2`

Dielectric constant k=2.1

Charge on parallel plate capacitor=
`Q=4.0\mu C=4* 10^(-6) C`

`1\mu C=10^(-6) C`

We know that

`V=(Qd)/(k\epsilon_0A)`

Where
`\epsilon_0=8.85* 10^(-12)F/m`

Substitute the values

`V=(4* 10^(-6)* 0.81* 10^(-3))/(2.1* 8.85* 10^(-12)* 1.3* 10^(-2)) V`

`V=1.34* 10^4 V`

Hence, the potential difference between the plates =
`1.34* 10^4 V`