A parallel plate capacitor has plates of area `200cm^2 and separation between the plates 1.00 mm. What potential difference will be developed if a charge of 1.00 nC )(i.e., 1.00X10^(-9) C) is given to the capacitor? If the plate separation is now increased to 2.00 mm, what will be the new potential difference?

Explanation

The capacitance of the capacitor is `
`C=(epsilon_o A)/(d)`
` `
`= 8.85 X 10^(_12) Fm ^(-1) X (200 X10^(-4) m^2)/(1X10^(-3) m)`
` `
`= 0.177 X 10 ^(-9) F = 0.177 nF. The potential difference between the plates is `
` V =(Q)/(C ) = (1nC)/(0.177 nF) = 5.65 volts. if the separation is increased from 1.00 mm to 2.00 mm, the capacitance is decreased by a factor of 2. If the charge remains the same, the potential difference will increase by a factor of 2. Thus, the new potential difference will be `
`5.65 volts X 2 = 11.3 volts.