Question

A parallel-plate capacitor has plates with an area of 0.012 m² and a separation of 0.009 m. The space between the plates is filled with a dielectric whose dielectric constant is 2. What is the potential difference between the plates when the charge on the capacitor plates is 5 ?C? (in kV)

Answer 1

We know that

The capacaitance of the capacitor with dielectric constant is given by

C =ke0A/d

We also know that Q =CV then V =Q/C

Given that

A parallel-plate capacitor has plates with an area of (A) = 0.012 m²

The separation between the plates is (d) = 0.009 m

Dielectric constant (k) =2

The  charge on the capacitor plates is (Q) = 5 uC =5*10^-6C

e0 =8.85*10^-12

Then the potenial diffrence is V =Q/C =Q/ke0A/d =5*10^-6C/(2)*8.85*10^-12)(0.012 m2)/0.009 m =0.211*10⁶V =211kV