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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance1 Mark
MCQ
The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?
  • A
    Charge on the capacitor.
  • B
    Potential difference across the capacitor.
  • C
    Energy of the capacitor.
  • ✓
    B and c both
✓

Answer

Correct option: D.
B and c both
Explanation:
Because the charge always remains conserved in an isolated system, it will remain the same.
Now,
$\text{V}=\frac{\text{Qd}}{\in_0\text{A}}$
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:
$\text{u}=\frac{1}{2}\in_0\text{E}^2$
So, u will remain constant with increase in distance between the plates.

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