- Charge on the capacitor.
- Potential difference across the capacitor.
- Energy of the capacitor.
- Energy density between the plates.
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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(a) |
(b) |
(c) |
(d) |
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(a) 100 watt lamp will fuse |
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