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

Question

asked May 20, 2021 46.4k views

1 Answer

Etherice · Answer 1 · 2021-05-24T06:31:13+0000

Answer:

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 =