# a parallel plate capacitor of capacitance C is charged to potential V by a battery without disconnecting the battery distance between the places tripled and dielectric medium of K = 10 is introduced between the plates of the capacitor explained giving reasons how is the following affected capacitance of a capacitor charge on capacitor and energy density of the capacitor

Given, the parallel plate capacitor has a capacitance C.

Let the plate area be A and the distance between the plates be d.

Then,

$C={\epsilon}_{o}\frac{A}{d}\phantom{\rule{0ex}{0ex}}$

If the distance between the plates is tripled and a dielectric medium of K=10 is introduced, the capacitance will change irrespective of the fact whether or not it is connected to a battery. The new capacitance will be C' which is given by:

$C\text{'}=K{\epsilon}_{o}\frac{A}{3d}=\frac{10}{3}\left({\epsilon}_{o}\frac{A}{d}\right)=\frac{10C}{3}\phantom{\rule{0ex}{0ex}}$

If earlier for a capacitance C, the charge across the plates is Q such that

$Q=CV\phantom{\rule{0ex}{0ex}}$

Then, the charge developed across the capacitance C' will be

$Q\text{'}=C\text{'}V=\frac{10C}{3}V=\frac{10Q}{3}\phantom{\rule{0ex}{0ex}}$

Ealier the energy density of the capacitor was

$E=\frac{1}{2}C{V}^{2}=\frac{{Q}^{2}}{2C}\phantom{\rule{0ex}{0ex}}$

After the change the energy density would be

$E\text{'}=\frac{1}{2}C\text{'}{V}^{2}=\frac{Q{\text{'}}^{2}}{2C\text{'}}\phantom{\rule{0ex}{0ex}}E\text{'}=\frac{1}{2}\times \frac{10}{3}C{V}^{2}=\frac{10}{3}E\phantom{\rule{0ex}{0ex}}$

Regards

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