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The separation between the plates of a isolated charged parallel plate capacitor is increased. Which of the following quantities will change?
charge on the capacitor
potential difference across the capacitor
energy of the capacitor
$B$ and $C$ both
Solution
(b) Potential difference across the capacitor
(c) Energy of the capacitor.
Because the charge always remains conserved in an isolated system, it will remain the same.
Now,
$V=\frac{Q d}{\epsilon_{0} A}$
Here, $\mathrm{Q}, \mathrm{A}$ and $\mathrm{d}$ are the charge, area and distance between the plates, respectively.
Thus, as d increases, V increases. Energy is given by $E=\frac{q V}{2}$
So , it will also increase.
Energy density u, that is, energy stored per unit volume in the electric field is given by $u=\frac{1}{2} \in_{0} E^{2}$
So, u will remain constant with increase in distance between the plates.
