A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from…

1. The potential difference between the plates decreases $K$ times.
2. The energy stored in the capacitor decreases $K$ times.
3. The change in energy stored is $\frac{1}{2}C{V}^2\left(\frac{1}{K}-1\right)$.
4. The charge on the capacitor is not conserved.

Solution

For air $K=1$,
$C_1=C$
$q=C_{1}V \dots \left(i\right)$.
The charge remains constant.
$U_1=\frac{q^2}{2C}$
$U_2=\frac{q^2}{2CK}$
Change in energy stored in the capacitor,
$U=U_2-U_1$
$=\frac{q^2}{2C} \left(\frac{1}{K}-1\right) =\frac{1}{2}C{V}^2 \left(\frac{1}{K}-1\right)$
$V\text{'}=\frac{q}{CK}\Rightarrow V\text{'}=\frac{V}{K}$
Here,1- Potential difference between the plates decreases by $K$ times.
3- The change in energy stored $∆U=\frac{1}{2} C{V}^2 \left(\frac{1}{K}-1\right)$.
4- Charge is conserved.
2- The energy stored in capacitor decrease by $K, U_2=\frac{1}{2}\frac{q^2}{CK}$.

Asked in: NEET 2015 (Phase 1)

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