Each plate of a parallel capacitor has area S = 5 × 10 –3 m 2 and are d = 8.85 mm apart. Plate A has a positive charge q 1 = 10 –10 Cb and plate B has charge q 2 = + 2 × 10 –10 Cb. Energy supplied by a battery of emf E = 10 volt when its positive terminal is connected with plate A and negative terminal with plate B is ….. 10 –9 Joule
Text Solution
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(0001)
Sol.
10 –10 – x = 2 × 10 –10 + x
2x = – 10 –10 x = –5 × 10
–11 Cb
Kirchoff law
q + x = CE
q = CE – x
= 
– (–5 × 10 –11 )
= 10 × 5 × 10 –12 + 5 × 10 –11 = 10 –10 Cb
Energy supplied
= qE = 10 –10 × 10 = 10 –9 Joule
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