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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance1 Mark
Question
The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?
✓

Answer

  1. Potential difference across the capacitor.
  2. Energy of the capacitor.
Explanation:
Because the charge always remains conserved in an isolated system, it will remain the same.
Now,
$\text{V}=\frac{\text{Qd}}{\epsilon_0\text{A}}$
Here, Q, A and d are the charge, area and distance between the plates, respectively.
Thus, as d increases, V increases.
Energy is given by:
$\text{E}=\frac{\text{qV}}{2}$ 
So, it will also increase.
Energy density u, that is, energy stored per unit volume in the electric field is given by:
$\text{u}=\frac{1}{2}\epsilon_0\text{E}^2$
So, u will remain constant with increase in distance between the plates.

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