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Manipal MET2015PhysicsCapacitance

A capacitor of capacitance 10 ~F is charged to potential 50 V with a battery. The battery is now disconnected and an additional charge 200 C is given to the positive plate of the capacitor. The potential difference across the capacitor will be

Options

  1. A50 V
  2. B80 V
  3. C100 V
  4. D60 V

Correct answer

D. 60 V

Step-by-step solution

Charge acquired by the plates of the capacitor q₀=C V=(10 ~F ) (50 ~V )=500 C Now, let the charge distribution is as follows. Total charge on positive plate has now become 700 C while that in negative plate is still -500 C , Here, charges are in C . Net electric field at point P is zero. array lc & (700-q) 2 A ₀ + q 2 A ₀ + ( 500-q 2 A ₀ )= q 2 A ₀ & q=600 C array Potential difference between the plates is V= q C = 600 C 10 ~F =60 ~V

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