Question

6. A parallel-plate capacitor has plates of area 500 cm2 and is connected across the terminals of a battery. After some time has passed, the capacitor is disconnected from the battery. When the plates are then moved 0.50 cm farther apart, the charge on each plate remains constant but the potential difference between the plates increases by 100 V. (a) What is the magnitude of the charge on each plate? (b) Determine the change in stored energy in the capacitor due to the movement of the plates.

Answer 1

6. plate area A = 500 cm² = 500x10m = 5x10m² Suppose plate separation is a 4 battery potential i E, so capacitance c = & A & charge over plates S = CE = EME Now New distance d'= d+0.50 = d70.005 d Va avea 205) = SME (8 +0.05) so V-E =100 = E($0.005) - € = 100 → £+ 0.005€ 8 = 100 => ļ the store a = 2x109 9 charge Q=&th = 8.854x100 x 5x10²x2x10 C Q = 8-854x10°C => [@=88549C

(6) Energy stored U= 2 av So change in energy Au=ZQV - 2085 = CV-F) = $48.854 X 16" X 100 J (AU= 4.427x10²5)