Question

An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

Answer 1

The new voltage between the plates of the capacitor is 18 V

Step-by-step explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

`C = (\epsilon _0A)/(d) \\\\q =CV = ((\epsilon _0A)/(d))V = (\epsilon _0A V)/(d)`

`q = (\epsilon _0A V)/(d)`

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

`q = (\epsilon _0A V)/(d) \\\\(\epsilon _0A V)/(d) = (\epsilon _0A V)/(d) \\\\(V_1)/(d_1) = (V_2)/(d_2) \\\\V_2 = (V_1d_2)/(d_1) \\\\(d_2 = 2d_1)\\\\V_2 = (V_1*2d_1)/(d_1) \\\\(V_1 = 9V)\\\\V_2 = (9*2d_1)/(d_1) \\\\V_2 = 9*2\\\\V_2 = 18 \ V`

Therefore, the new voltage between the plates of the capacitor is 18 V