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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance3 Marks
Question
Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C' and having charge 'Q'.
How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant 'K'?
✓

Answer

Potential difference between the plates of capacitor
V = q/C
Work done to add additional charge dq on the capacitor
dw = V x dq= (q/C) x dq
 $\therefore$Total energy stored in the capacitor
$\text{U} = \int\text{dw} = \int_{0}^{Q}\frac{\text{q}}{\text{c}}\text{dq} = \frac{1}{2}\frac{\text{Q}^{2}}{\text{c}}$
When battery is disconnected:
  1. Energy stored will be decreased or energy stored = 1/K times the initial energy.
  2. Electric field would decrease or E’ = E/K.
Alternate Answer
When battery is connected:
  1. Energy stored will increase or become K times the initial energy.
  2. Electric field will not change.

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