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The work done in placing the dielectric slab inside one of the capacitors as shown in the diagram.

Options

  1. AC V 2 2 K – 1 K + 1
  2. BC V 2 4 K – 1 K + 1
  3. CC V 2 4 K + 1 K – 1
  4. DC V 2 2 K + 1 K – 1

Correct answer

B. C V 2 4 K – 1 K + 1

Step-by-step solution

U i = 1 2 C / 2 V 2 = C V 2 4 U f = 1 2 C × K C C + K C V 2 = K C V 2 2 1 + K W = Δ U = U f – U i = K C V 2 2 1 + K - C V 2 4 = C V 2 2 K 1 + K – 1 2 = C V 2 2 2 K – 1 – K 2 1 + K = C V 2 4 K – 1 K + 1

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