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A parallel plate capacitor with plate area A and plate separation d = 2 m has a capacitance of 4 μF . The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K = 3 (as shown in figure) will be

Options

  1. A2   μF
  2. B32   μF
  3. C6   μF
  4. D8   μF

Correct answer

C. 6   μF

Step-by-step solution

This parallel plate capacitor can be divided into two capacitors. One with dielectric C 1 and other without dielectric C 2 . The two capacitors will be in series. Initially C = ε 0 A d = 4   μF Now, C 1 = k ε 0 A d 2 = 2 × 3 × ε 0 A d = 24   μF and C 2 = ε 0 A d 2 = 2 × ε 0 A d = 8   μF Finally C ' = C 1 C 2 C 1 + C 2 = 24 × 8 24 + 8 = 6   μF

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