Two dielectrics of dielectric constants K1 and K2 are filled in gap of parallel plate capacitor as shown in figure The capacitance of capacitor will be:

Question

Two dielectrics of dielectric constants K1 and K2 are filled in gap of parallel plate capacitor as shown in figure The capacitance of capacitor will be: (A) (ε0 A(K1+K2)/2 d) (B) (ε0 A/2 d)((K1+K2/K1 K2)) (C) (ε0/d)((K1 K2/K1+K2)) (D) (ε0 A/d)((K1+K2/K1 K2))

Tardigrade Admin · Accepted Answer

In each capacitor, the area of the plate will be

\frac{A}{2} .

In the given combination, the arrangement is equivalent to two capacitors connected in parallel. Also in each capacitor the area of the plate will be

\frac{A}{2} .

Therefore, equivalent capacitance

C^{'} = C_{1} + C_{2}

= \frac{K _{1} ε _{0} A /2}{d} + \frac{K _{2} ε _{0} A /2}{d}

= \frac{ε _{0} A}{2 d} (K_{1} + K_{2})

Q. Two dielectrics of dielectric constants $K_{1}$ and $K_{2}$ are filled in gap of parallel plate capacitor as shown in figure
The capacitance of capacitor will be:

$\frac{ε _{0} A ( K _{1} + K _{2} )}{2 d}$

$\frac{ε _{0} A}{2 d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$

$\frac{ε _{0}}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

$\frac{ε _{0} A}{d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$

Q. Two dielectrics of dielectric constants K1​ and K2​ are filled in gap of parallel plate capacitor as shown in figureThe capacitance of capacitor will be:

2dε0​A(K1​+K2​)​

2dε0​A​(K1​K2​K1​+K2​​)

dε0​​(K1​+K2​K1​K2​​)

dε0​A​(K1​K2​K1​+K2​​)

Q. Two dielectrics of dielectric constants $K_{1}$ and $K_{2}$ are filled in gap of parallel plate capacitor as shown in figure
The capacitance of capacitor will be:

$\frac{ε _{0} A ( K _{1} + K _{2} )}{2 d}$

$\frac{ε _{0} A}{2 d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$

$\frac{ε _{0}}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

$\frac{ε _{0} A}{d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$