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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Bihar 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.
  • D
    Energy density between the plates.
✓

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,

$\text{V}=\frac{\text{Qd}}{\in_0\text{A}}$

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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