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JEE Main2016PhysicsCapacitanceActual

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μ F and 9 μ F capacitors), at a point distant 30 m from it, would equal:

Options

  1. A420 N/C
  2. B480 N/C
  3. C240 N/C
  4. D360 N/C

Correct answer

A. 420 N/C

Step-by-step solution

C e q = 5 μ F Potential of 4 μF = 6 volt ∴ charge on 4 μF q 4 = 24 μC Potential of 9 μF = 2 volt ∴ charge on 9 μF q 9 = 18 μC Total charge ( q ) = 42 μC E = k q r 2 = 9 × 10 9 × 42 × 10 - 6 900 = 420 N / C

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