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Two identical metal plates are given positive charges Q 1 and Q 2 ( < Q 1 ) respectively. If they are now brought close together to form a parallel plates capacitor with capacitance C , the potential difference between them is

Options

  1. AQ 1 + Q 2 2 C
  2. BQ 1 + Q 2 C
  3. CQ 1 - Q 2 C
  4. DQ 1 - Q 2 2 C

Correct answer

D. Q 1 - Q 2 2 C

Step-by-step solution

For charge Q 1 , electric field is E 1 = Q 1 2 ε 0 A where A is area. For charge Q 2 electric field is E 2 = Q 2 2 ε 0 A Resultant electric field E = E 1 - E 2 = ( Q 1 - Q 2 ) 2 ε 0 A Potential difference between plates when they are brought close together to form a parallel plate capacitor is V = E d = ( Q 1 - Q 2 ) 2 ε 0 A d Since, C = ε 0 A d for parallel plate capacitor. V = ( Q 1 - Q 2 ) 2 C

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