Question

A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness (3/4)d, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:

(Given $C_0$ = capacitance of capacitor with air as medium between plates.)

• A. $\frac{4KC_0}{3+K}$
• B. $\frac{3KC_0}{3+K}$
• C. $\frac{3+K}{4KC_0}$
• D. $\frac{K}{4+K}$

Answer 1

Answer: (A)

$\frac{4KC_0}{3+K}$

Answer 2

$x + y +$ $\frac{3d}{4}$ $= dx + y =$ $\frac{d}{4}$

$\frac{A∈_0}d = C_0$

$∆V = Ex +\frac{ E}{K} × \frac{3d}{4} + Ey= \frac{3Ed}{4K }+ E(x + y)$

$∆V = E[\frac{3d}{4K }+ \frac{d}{4}]$

$∆V = \frac{σ}{∈_0}[\frac{3d + dK}{4K}] = \frac{Qd}{A∈_0}[\frac{3 + K}{4K}]$

$\frac{Q}{∆V} = C =\frac{ A∈_0}d[\frac{4K}{K + 3}]$

$C=\frac{C_04K}{3+K}$

∴ $C=\frac{4KC_0}{3+K}$