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An uncharged parallel plate capacitor having a dielectric of dielectric constant K is connected to a similar air-filled capacitor charged to a potential V. The two share the charge, and the common potential is $V’$. The dielectric constant K is:

Last updated date: 22nd Jul 2024
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Hint:We will use the concept of common potential. The common potential is utilized in the capacitor. The significance of common potential is that meanwhile, two capacitors are charged to various potentials connected by a wirework, then both have varied potentials. The charge moves from higher to lower potential. This flow lasts till the charge does not match. A time will happen when both capacitors have an equal charge as well as equal potential. This is called common potential.

Complete step-by-step solution:
Air-filled capacitor having capacitance, $C_{1} = C$.
Dielectric filled capacitor having capacitance, $C_{2} = KC$.
Air-filled capacitor having charge, $Q = CV$.
Now, the charge is shared by air-filled capacitor and dielectric filled capacitor. Then, charge is divided between both equally.
Charge on air filled capacitor, $Q_{1} = \dfrac{CV}{2}$
Charge on dielectric filled capacitor, $Q_{2} = \dfrac{CV}{2}$.
Common Potential is given by:
$V’ = \dfrac{Q_{1} + Q_{2}}{ C_{1} + C_{2}}$
$\implies V’ = \dfrac{\dfrac{CV}{2}+\dfrac{CV}{2}}{ C +KC}$
$\implies V’ = \dfrac{CV}{ C +KC}$
$\implies V’ = \dfrac{V}{ 1 +K}$
$\implies 1 +K = \dfrac{V}{ V’ }$
$\implies K = \dfrac{V}{ V’ } -1$
This is dielectric constant.

Note: In the uncharged state, the charge on each one of the conductors in the capacitor is zero. A charge Q is transferred from one conductor to another during the charging process, giving one conductor a charge +Q and the other a charge. A potential difference is generated, with the positively charged conductor greater than the negatively charged conductor. Remark that whether charged or uncharged, the total charge on the capacitor as a sum is zero.