A capacitor has some dielectric between its plates and the capacitor is connected to a $\mathrm{D.C.}$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.
A capacitor has some dielectric between its plates and the capacitor is connected to a $\mathrm{D.C.}$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.
Capacitance of capacitor with dielectric constant $K$,
$\mathrm{C}=\frac{\mathrm{K} \in_{0} \mathrm{~A}}{d}$
where $K$ is positive and its value is greater then $1$ . So by removing dielectric slab and keeping $A$ and $d$ constant capacitance $\mathrm{C}$ of capacitor will decrease. When battery and dielectric slab from capacitor is removed the charge remains the same.
As the energy in capacitor $U=\frac{q^{2}}{2 C}$, when capacitance $C$ is decreased by removing dielectric slab but $q$ is the same hence in $\mathrm{U} \propto \frac{1}{\mathrm{C}} \mathrm{C}$ is decreased so $\mathrm{U}$ will increase.
By equation $\mathrm{C}=\frac{q \mathrm{Q}}{\mathrm{V}}, q$ is constant and $\mathrm{C}$ decreases and $\mathrm{V}$ should increase.
Now $d$ is constant and $\mathrm{V}$ increases hence according to $\mathrm{E}=\frac{\mathrm{V}}{d} \mathrm{E}$ is increases.
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