Question

The capacity of a parallel plate condenser is $C _ { 0 }$ . If a dielectric of relative permitivity$\varepsilon _ { r }$ and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of $\frac { C } { C _ { 0 } }$ will be

• A. $\frac { 5 \varepsilon _ { r } } { 4 \varepsilon _ { r } + 1 }$
• B. $\frac { 4 \varepsilon _ { r } } { 3 \varepsilon _ { r } + 1 }$
• C. $\frac { 3 \varepsilon _ { r } } { 2 \varepsilon _ { r } + 1 }$
• D. $\frac { 2 \varepsilon _ { r } } { \varepsilon _ { r } + 1 }$

Answer 1

Answer: (B)

$\frac { 4 \varepsilon _ { r } } { 3 \varepsilon _ { r } + 1 }$

Answer 2

Initially capacitance $C _ { 0 } = \frac { \varepsilon _ { 0 } A } { d }$ ……..(i) Finally

capacitance $C = \frac { \varepsilon _ { 0 } A } { d - \frac { d } { 4 } + \frac { d / 4 } { \varepsilon _ { r } } }$ ……..(ii)

By dividing equation (ii) by equation (i) $\frac { C } { C _ { 0 } } = \frac { 4 \varepsilon _ { r } } { 3 \varepsilon _ { r } + 1 }$