Question

A parallel-plate capacitor has plates with an area of 0.012
m^{2} 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 m^{2}

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^{-6}C

e0 =8.85*10^{-12}

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

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