Question

A capacitor is formed by two square metal-plate of edge a, separated by a distance d. Dielectric of dielectric constant K₁& K₂ are filled in the gap as shown in the fig. Find the capacitance of the system-

• A. $\frac{\varepsilon_{0}a^{2}K_{1}K_{2}}{(K_{1}–K_{2})d}$ln $\frac{K_{1}}{K_{2}}$
• B. $\frac{\varepsilon_{0}a^{2}K_{1}K_{2}}{(K_{2}–K_{1})d}$ln $\frac{K_{2}}{K_{1}}$
• C. $\frac{\varepsilon_{0}a^{2}K_{1}}{(K_{1}–K_{2})d}$ ln $\frac{K_{1}}{K_{2}}$
• D. $\frac{\varepsilon_{0}a^{2}K_{2}}{(K_{2}–K_{1})d}$ln $\frac{K_{2}}{K_{1}}$

Answer 1

Answer: (A)

$\frac{\varepsilon_{0}a^{2}K_{1}K_{2}}{(K_{1}–K_{2})d}$ln $\frac{K_{1}}{K_{2}}$

Answer 2

Q = ̃ y = $\frac { \mathrm { xd } } { \mathrm { a } }$

dC₁ = $\frac { \varepsilon _ { 0 } \mathrm {~K} _ { 1 } ( \mathrm { adx } ) } { \mathrm { d } - \frac { \mathrm { xd } } { \mathrm { a } } }$ , dC₂ =

\ = $\frac { 1 } { \mathrm { dC } _ { 1 } }$ + $\frac { 1 } { \mathrm { dC } _ { 2 } }$

= +

=

= $\left[ \frac { \mathrm { K } _ { 2 } + \mathrm { x } / \mathrm { a } \left( \mathrm { K } _ { 1 } - \mathrm { K } _ { 2 } \right) } { \mathrm { K } _ { 1 } \mathrm {~K} _ { 2 } } \right]$

dC =

\ C==

C = log (K₁/K₂)