What is the potential difference between parallel plate capacitor plates, and what happens to potential difference if plate separation is doubled?A parallel plate capacitor has a capacitance of 1.2...

What is the potential difference between parallel plate capacitor plates, and what happens to potential difference if plate separation is doubled?

A parallel plate capacitor has a capacitance of 1.2 nF. There is a charge of magnitude 0.800 microcoulombs on each plate. When plate separation is doubled, charge is kept constant.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since the values of capacitors are specified in farads, the first step is to convert nanofarads into farads:

C = 1.20*10^-9 F

Now, we'll convert the microCoulombs in Coulombs:

Q = 0.800*10^-6 C

Now, we'll write the potential difference for a parallel plate capacitor:

V = Q/C

We'll substitute the values for charge and capacitance and we'll get:

V = 0.800*10^-6/1.20*10^-9

V = 0.667*10^(9-6) V

V = 0.667*10^(3) V

We know that the capacitance of parallel plate capacitor increases with area and it decreases with separation d:

C = A*e0/d

e0 = permittivity of dielectric

d = width of dielectric (plate separation)

If d is doubled, we'llĀ get a double decreased capacitance:

C = A*e0/2d

If the charge is kept constant:

V = Q/C

where C is halved, so the potential difference is doubled when d is doubled.