When a capacitor has a charge of magnitude 80 μC on each plate the potential difference across the plates is 16 V. How much energy is stored in this capacitor when the potential difference across its plates is 42 V?

  4.410 mJ

Explanation:

The capacitance is ...

  C = Q/V

  C = (80 μC)/(16 V) = 5 μF . . . . . variable C = capacitance; unit C = coulombs

Then the energy stored at 42 V is ...

  E = 1/2CV² = (1/2)(5 μF)(42 V)² = 4.410 mJ

The 5 μF capacitor stores about 4.41 millijoules at 42 V.

lexy is right because she has the labeled s

f=ma newton's second law 7x8=56n