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Question

Two parallel plate capacitor of capacitances C and 2C are connected in parallel and charged to a potential difference V . If the battery
is disconnected and the space between the plate of the capacitor of capacince C is completely filled with a material of dielectric constant K, then the potential difference across the capacitor will be come


  1. 3V(K+2)
  2. (K+2)3V
  3. 3V(K+2)
  4. 3(K+2)V

A
3V(K+2)
B
(K+2)3V
C
3V(K+2)
D
3(K+2)V
Solution
Verified by Toppr

The original capacitance of the parallel combination of C and 2 C is = C +2C = 3C
Total charge Q = 3CV
When the capacitor C is filled with a dielectric, its capacitance becomes KC.
Therefore, after the capacitor is filled with dielectric the capacitance of the combination, C' = KC + 2C = C (K + 2).
However, the charge remains the same, i.e. Q = 3CV
Therefore, the potential difference across the capacitors will be
begin mathsize 12px style fraction numerator Q over denominator C apostrophe end fraction equals fraction numerator 3 C V over denominator open parentheses K space plus space 2 close parentheses begin display style C end style end fraction
therefore fraction numerator begin display style Q end style over denominator begin display style C apostrophe end style end fraction equals fraction numerator begin display style 3 V end style over denominator begin display style open parentheses K space plus space 2 close parentheses end style end fraction end style

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