An ideal parallel - plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference across the plates. If the separation between the plates is now doubled, the amount of electrical energy stored on the capacitor will be cut in half. double. not change. quadruple. be cut in fourth.

The capacitance is cut in half.

Explanation:

The capacitance of a plate capacitor is directly proportional to the area A of the plates and inversely proportional to the distance between the plates d. So if the distance was doubled we should expect that the capacitance would be cut in half. That can be verified by the following equation that is used to compute the capacitance in such cases:

C = (\epsilon)*(A/d)

Where \epsilon is a constant that represents the characteristics for the insulator between the plates. A is the area of the plates and d is the distance between them. When we double d we have a new capacitance, given by:

C_new = (\epsilon)*(A/2d)

C_new = (1/2)*[(\epsilon)*(A/d)]

Since C = (\epsilon)*(A/d)] we have:

C_new = (1/2)*C

there are no contact forces mentioned on the list of choices that you've provided.

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