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ozzi
3 years ago
12

Addition of a metal slab of thickness "a" between the plates of a parallel plate capacitor of plate separation "d" is equivalent

to introducing a dielectric with dielectric constant "K" between the plates. Find an expression for "K" in terms of "a" and "d"?
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

K =  \frac{d}{d+a}

Explanation:

The capacitance of a capacitor in terms of the dielectric constant, area of the plate and the distance separating the plate is given by:

C = \frac{\epsilon A}{d}

Where A = Area of the plate

d = distance between the plates

\epsilon = dielectric constant

Case 1:

When a meta slab of thickness, a, is added between the plates of the parallel plate capacitor , the effective separation between the plates becomes d+a

Therefore the capacitance of the capacitor becomes:

  C = \frac{\epsilon A}{d + a} .......................(1)

Case 2:

Introducing a dielectric with dielectric constant K between the plates, the capacitance of the capacitor becomes:

C = \frac{K\epsilon A}{d}.........................(2)

Equating (1) and (2)

\frac{K\epsilon A}{d} = \frac{\epsilon A}{d+a}\\\frac{K}{d} = \frac{1}{d+a} \\K = \frac{d}{d+a}

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