# 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

Dear Student,
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
$E\text{'}=\frac{1}{2}C\text{'}{V}^{2}=\frac{Q{\text{'}}^{2}}{2C\text{'}}\phantom{\rule{0ex}{0ex}}E\text{'}=\frac{1}{2}×\frac{10}{3}C{V}^{2}=\frac{10}{3}E\phantom{\rule{0ex}{0ex}}$
Regards

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