Question

Capacitance of a capacitor becomes 7/6 times its original value if a dielectric slab of thickness t = (2/3)d is introduced between plates. d is the separation between plates. The dielectric constant of dielectric slab is-

• A. $\frac{14}{11}$
• B. $\frac{11}{14}$
• C. $\frac{7}{11}$
• D. $\frac{11}{7}$

Answer 1

Answer: (A)

$\frac{14}{11}$

Answer 2

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

Ž C' = Ž $\frac { 7 } { 6 }$ C = $\frac { \varepsilon _ { 0 } \mathrm {~A} } { \frac { \mathrm { d } } { 3 } + \frac { 2 \mathrm {~d} } { 3 \mathrm { k } } }$

Ž $\frac { 7 } { 6 }$ C = $\frac { \varepsilon _ { 0 } \mathrm {~A} } { \mathrm {~d} } \left( \frac { 3 \mathrm { k } } { \mathrm { k } + 2 } \right)$ Ž $\frac { 7 } { 6 }$ C = C $\left( \frac { 3 \mathrm { k } } { \mathrm { k } + 2 } \right)$

7k + 14 = 18 k Ž k = $\frac { 14 } { 11 }$