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KCET2013PhysicsCapacitance

See the diagram. Area of each plate is 2.0 ~m ² and d=2 10⁻³ ~m . A charge of 8.85 10⁻⁸ C is given to Q . Then the potential of Q becomes

Options

  1. A13 ~V
  2. B10 ~V
  3. C6.67 ~V
  4. D8.825 ~V

Correct answer

C. 6.67 ~V

Step-by-step solution

In the given arrangement, plate Q is common for two capacitors which are connected in parallel. aligned & C_ eff =C_ P =C₁+C₂ & C_ P = ₀ A d + ₀ A 2 d = 2 ₀ A 2 d aligned The given arrangement can be shown as Let V be the potential difference across the capacitor, which is equal to the potential of the plate Q So, C= q V i.e., C_ eff = q V i.e., 3 ₀ A 2 d = 8.85 10⁻⁸ V aligned V &= 8.85 10⁻⁸ 2 2 10⁻³ 3 8.85 10⁻¹² 2 &= 2 3 10= 20 3 =6.67 ~V aligned

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