Question

A dielectric of constant 6 is to be filled in the gap between plates of a parallel plate capacitors. What is the factor by which the gap should be changed to maintain the same capacitance

Answer 1

The gap between plate of capacitor should be changed by factor 6 to maintain the capacitance.

Explanation:

Capacitance of capacitor C = eA/d

• Where, e = permittivity of air
• A = area of plate
• d = distance between two plate

Effect of dielectric:

Suppose, A dielectric of permittivity e is to be filled in the gap.

New capacitance = KeA/d

Because, Relative permittivity = Dielectric constant (K) = e'/e

Thus, e' = K.e

Calculation:

A) C1 = ęA/d1

B) C2 = KęA/d2

Here, C1 = C2 and Area A is constant

=> So, e/d1 = Ke/d2

=> K=6

=> d1 = d2/6

=> d2 = 6 x d1

Thus the gap between plate of capacitor should be changed by factor 6 to maintain the capacitance.