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Byju's Answer

Standard XII

Physics

Capacitance of a Random Conductor

Two materials...

Question

Two materials of dielectric constants $k_{1} a n d k_{2}$ are introduced to fill the space between the two parallel plates of a capacitor as shown in the figure. The capacitance of the capacitor is :

\frac{A \in_{0} (k_{1} + k_{2})}{2 d}

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\frac{2 A \in_{0}}{d} (\frac{k_{1} k_{2}}{k_{1} + k_{2}})

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\frac{A \in_{0}}{d} (\frac{k_{1} k_{2}}{k_{1} + k_{2}})

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\frac{A \in_{0} (k_{1} + k_{2})}{2 d k_{1} k_{2}}

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Solution

The correct option is A $\frac{A \in_{0} (k_{1} + k_{2})}{2 d}$
$C = K \frac{E_{0} A}{d}$
$C = C_{1} + C_{2} = \frac{E_{0} \times \frac{A}{2} \times K_{1}}{d} + \frac{E_{0} \times \frac{A}{2} \times K_{2}}{d} = \frac{E_{0} A (K_{1} + K_{2})}{2 d}$

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Two materials of dielectric constants k1andk2 are introduced to fill the space between the two parallel plates of a capacitor as shown in the figure. The capacitance of the capacitor is :

The correct option is A A∈0(k1+k2)2dC=KE0AdC=C1+C2=E0×A2×K1d+E0×A2×K2d=E0A(K1+K2)2d

Two materials of dielectric constants $k_{1} a n d k_{2}$ are introduced to fill the space between the two parallel plates of a capacitor as shown in the figure. The capacitance of the capacitor is :

The correct option is A $\frac{A \in_{0} (k_{1} + k_{2})}{2 d}$
$C = K \frac{E_{0} A}{d}$
$C = C_{1} + C_{2} = \frac{E_{0} \times \frac{A}{2} \times K_{1}}{d} + \frac{E_{0} \times \frac{A}{2} \times K_{2}}{d} = \frac{E_{0} A (K_{1} + K_{2})}{2 d}$