A slab of material of dielectric constant K has same area as plates of a parallel-plate capacitor bu

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2. Electric Potential and Capacitance

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A slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$, where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

Option A

Option B

Option C

Option D

Solution

Let $E_{0}=V_{0} / d$ be the electric field between the plates when there is no dielectric and the potential difference is $V_{0} .$ If the dielectric is now inserted, the electric field in the dielectric will be $E=E_{0} / K$ The potential difference will then be

$V=E_{0}\left(\frac{1}{4} d\right)+\frac{E_{0}}{K}\left(\frac{3}{4} d\right)$

$=E_{0} d\left(\frac{1}{4}+\frac{3}{4 K}\right)=V_{0} \frac{K+3}{4 K}$

The potential difference decreases by the factor $(K+3) / K$ while the free charge $Q_{0}$ on the plates remains unchanged. The capacitance thus increases

$C=\frac{Q_{0}}{V}=\frac{4 K}{K+3} \frac{Q_{0}}{V_{0}}=\frac{4 K}{K+3} C_{0}$

Standard 12

Physics

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A slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$, where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

Solution

Similar Questions

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