An empty parallel plate capacitor is connected between the terminals of a 10.1-V battery and charges 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

V' = 20.2 V

Explanation:

The formula for the capacitance of a parallel plate capacitor, with air as a medium between the plates, is given by the following formula:

C = A∈₀/d

where,

V = Capacitance of capacitor

A = Area of plate

∈₀ = permittivity pf free space

d = distance between plates

Keeping all the other variables constant, the relationship between capacitance and the distance between plates is given as:

C α 1/d

Therefore, when the distance between plates is doubled, it implies that the capacitance of the capacitor becomes half. That is:

C' = C/2

Now the formula for capacitance of a capacitor, is also given as:

Q = CV

C = Q/V

where,

Q = charge stored in capacitor

V = Voltage across capacitor

Keeping the charge constant the relationship between capacitance and voltage is:

C α 1/V

Therefore, when the capacitance is halved, the voltage must be doubled, keeping the charge constant:

V' = 2V = (2)(10.1 V)

V' = 20.2 V

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