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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance3 Marks
Question
A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected? Justify your answer.
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Answer

Original capacitance $\text{C}_{o} = \frac{\text{Q}}{\text{V}_{o}} = \varepsilon_{0}\frac{\it{\text{A}}}{\text{d}}$ When a dielectric is inserted:
  1. Capacitance
$\bigg( = \text{K}\in_{o}\frac{\it{\text{A}}}{\text{d}}\bigg)\text{ increases}$
  1. Electric Field.
$ = \bigg(\frac{\sigma - \sigma\text{P}}{\in_{o}}\bigg)\text{decreases}$
  1. Energy stored
$\bigg(\text{W} = \frac{1}{2}\text{KC}_{o}\cdot\frac{\text{Q}^{2}}{\text{C}_{o}\text{K}^{2}} =\frac{1}{2}\frac{\text{Q}^{2}}{\text{C}_{o}}\cdot\frac{1}{\text{k}}\bigg)\text{decrease}.$

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