Question

A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be.

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Answer 2

: $\mathrm { C } = \frac { \varepsilon _ { 0 } \mathrm {~A} } { \mathrm {~d} }$

When it is half filled with dielectric of K = 5, its capacitance becomes which is series combination of

and $\mathrm { C } _ { 2 } = \frac { \varepsilon _ { 0 } \mathrm {~A} } { \mathrm {~d} / 2 }$

$\Rightarrow \mathrm { C } _ { \mathrm { s } } = \frac { 5 } { 3 } \frac { \varepsilon _ { 0 } \mathrm {~A} } { \mathrm {~d} } = \frac { 5 \mathrm { C } } { 3 }$ (using (i))

Percentage increase in capacitance

$= \frac { \frac { 5 } { 3 } \mathrm { C } - \mathrm { C } } { \mathrm { C } } \times 100 = \frac { 2 } { 3 } \times 100 = 6.6 \%$