Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery. A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected, and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected:
- The electric field between the plates of the capacitor.
- The energy stored in the capacitor.
CBSE OUTSIDE DELHI - SET 3 PANCHKULA 2015
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Process of charging:
The electrons, from the plate of the capacitor, which is connected to the positive terminal of the battery, move towards the battery. The reverse happens at the other plate. Hence, the plates get positively and negatively charged respectively.
Effect of dielectric:
$\text{E} = \frac{\text{E}_{1}}{\text{K}}$
$ = \frac{\text{K}}{2}\text{C}_{\circ} , $
$\therefore\text{C} < \text{C}_{\circ}$
Initially Energy $ = \frac{\text{Q}^{2}}{2\text{c}}$ and Energy $ = \frac{\text{Q}^{2}}{\text{c}}.\frac{2}{\text{k}}\text{as }1 < \text{K} < 2.$
The electrons, from the plate of the capacitor, which is connected to the positive terminal of the battery, move towards the battery. The reverse happens at the other plate. Hence, the plates get positively and negatively charged respectively.
Effect of dielectric:
- Electric fields decreases Justification.
$\text{E} = \frac{\text{E}_{1}}{\text{K}}$
- Energy stored increases:
$ = \frac{\text{K}}{2}\text{C}_{\circ} , $
$\therefore\text{C} < \text{C}_{\circ}$
Initially Energy $ = \frac{\text{Q}^{2}}{2\text{c}}$ and Energy $ = \frac{\text{Q}^{2}}{\text{c}}.\frac{2}{\text{k}}\text{as }1 < \text{K} < 2.$
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