An uncharged parallel plate capacitor filled with a dielectric of dielectric constant K is connected to an air filled identical parallel capacitor charged to potential $V_{1}$ . If the common potential is $V_{2}$ , the value of K is

\frac{V_{1} - V_{2}}{V_{1}}

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\frac{V_{1}}{V_{1} - V_{2}}

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\frac{V_{2}}{V_{1} - V_{2}}

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\frac{V_{1} - V_{2}}{V_{2}}

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\frac{V_{1} - V_{2}}{V_{1} + V_{2}}

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Solution

The correct option is D $\frac{V_{1} - V_{2}}{V_{2}}$
If the capacitance of the air filled capacitor is $C_{1} = C$ , then the capacitance of the dielectric field capacitor will be $C_{2} = K C$
Now, the common potential is $V_{2} = \frac{Q_{1} + Q_{2}}{C_{1} + C_{2}}$
Here, $Q_{1} = C_{1} V_{1}$ , charge for air filled capacitor and as the dielectric filled capacitor is uncharged, $Q_{2} = 0$
Thus, $V_{2} = \frac{C_{1} V_{1} + 0}{C_{1} + C_{2}}$
or $V_{2} = \frac{C V_{1}}{C + K C}$
or $K + 1 = V_{1} / V_{2}$
or $K = \frac{V_{1} - V_{2}}{V_{2}}$

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An uncharged parallel plate capacitor filled with a dielectric of dielectric constant K is connected to an air filled identical parallel capacitor charged to potential V1. If the common potential is V2, the value of K is

An uncharged parallel plate capacitor filled with a dielectric of dielectric constant K is connected to an air filled identical parallel capacitor charged to potential $V_{1}$ . If the common potential is $V_{2}$ , the value of K is