   Chapter 16, Problem 51P

Chapter
Section
Textbook Problem

A parallel-plate capacitor has capacitance 3.00 μF. (a) How much energy is stored in the capacitor if it is connected to a 6.00-V battery? (b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? (c) The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored? (Answer each part in microjoules.)

(a)

To determine
The energy stored in the capacitor.

Explanation

Given info: The capacitance is 3.00μF . The potential difference applied by the battery is 6.00 V.

Explanation:

Formula to calculate the energy stored is,

E=12C(ΔV)2

• C is the capacitance.
• ΔV is the potential difference.

Substitute 3.00μF for C and 6.00 V for ΔV

E=12(3

(b)

To determine
The energy stored in the capacitor.

(c)

To determine
The energy stored in the capacitor when the battery is reconnected.

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