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AP EAMCET202120 Aug 2021Morning ShiftPhysicsCapacitanceActual

A 60 μF parallel plate capacitor whose plates are separated by 6 mm is charged to 250 V , and then the charging source is removed. When a slab of dielectric constant 5 and thickness 3 mm is placed between the plates, find the change in the potential difference across the capacitor?

Options

  1. A250   V
  2. B100   V
  3. C150   V
  4. D75   V

Correct answer

B. 100   V

Step-by-step solution

Capacitance of capacitor after inserting a slab of dielectric constant 5   and of thickness t = 3   mm is, C ' = C 1 1 - t d + t k d ⇒ C ' = 12   μF 1 1 - 3   mm 6   mm + 3   mm 5 × 6   mm = 12   μF × 20 12 ⇒ C ' = 20   μF Now, after removing the charging source from the capacitor the charge remain same, q = C V = C ' V ' ⇒ C V = C ' V ' ⇒ 12   μF × 250   V = 20   μ

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