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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance5 Marks
Question
A capacitor has some dielectric between its plates, and the capacitor is connected to a DC 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.
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Answer

The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant K is given by $\text{C}=\frac{\text{K}\in_0\text{A}}{\text{d}}$.
The capacitance of the parallel plate capacitor decreases with the removal of dielectric medium as for air or vacuum K = 1 and for dielectric K > 1.
If we disconnect the battery from capacitor, then the charge stored will remain the same due to conservation of charge.
The energy stored in an isolated charge capacitor $\text{U}=\frac{\text{q}^2}{2\text{C}}$ as q is constant, energy stored $\text{U}\propto\frac{1}{\text{C}}$. As C decreases with the removal of dielectric medium, therefore energy stored increases.
The potential difference across the plates of the capacitor is given by $\text{V}=\frac{\text{q}}{\text{C}}$.
Since q is constant and C decreases which in turn increases V and therefore E increases as $\text{E}=\frac{\text{V}}{\text{d}}$.
Quantity
Capacity C' = KC C' = KC
Charge Q' = Q (Charge is conserved) Q' = KQ
Potential
$\text{V}'=\frac{\text{V}}{\text{K}}$
 V' = V (Since Battery maintains the potential difference)
Intensity
$\text{E}'=\frac{\text{E}}{\text{K}}$
 E' = E
Energy
$\text{U}'=\frac{\text{U}}{\text{K}}$
 U' = UK

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