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Concepts Of Physics MCQ Edition [Volume 2]PhysicsCapacitors

A parallel-plate capacitor with a capacitance of 5 F is connected to a battery having an emf of 6 V . The separation between the plates is 2 mm . Calculate the electric field between the plates.

Options

  1. A1.5 10^3 V m ⁻¹
  2. B6 10^3 V m ⁻¹
  3. C3 10^3 V m ⁻¹
  4. D1.2 10^4 V m ⁻¹

Correct answer

C. 3 10^3 V m ⁻¹

Step-by-step solution

Given, potential difference across the plates, V = 6 V Separation between the plates, d = 2 mm = 2 10⁻³ m The electric field E between the plates of a parallel-plate capacitor is given by the relation: E = V d Substituting the given values: E = 6 2 10⁻³ E = 3 10^3 V m ⁻¹ The given value of capacitance is extra information and is not required for this calculation.

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