Multiple choice

A parallel-plate air capacitor of area 25 cm2 and with plates 1 mm apart is charged to a potential of 100 V. Calculate the new energy by separating the new plates by 0.002 m?

  1. 2.2 x 10-7 J

  2. 1.81 x 107 J

  3. 0.55 x 10-7 J

  4. 0.45 x 107 J

  5. 1.21 x 10-14 J
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Energy = ½ CV2 = ½Є0AV2/d = [8.85 x 10-12 x 25 x 10-4 x 104] /2 x 0.001 = 1.1 x 10-7J New plate separation = 0.002 m; potential across plates is still 100 V. New energy = 0.5 x original energy = 0.55 x 10-7 J The difference in energy is explained by the movement of charge in the wires as the capacitor partly discharges to maintain the potential.

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