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A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K which can just fill the air gap of the capacitor is now inserted in it. Which of the following is incorrect?
A
The potential difference between the plates decreases K times
B
The energy stored in the capacitor decreases K times
C
The change in energy stored is $\frac{1}{2}CV^2\left(\frac{1}{K} - 1\right)$
D
The charge on the capacitor is not conserved
Detailed Solution
On charging, $Q = CV$. After disconnecting, the charge cannot change, so the charge is conserved — the statement that it is not conserved is incorrect.
With the dielectric: $C' = KC$ and $V' = \frac{V}{K}$ (potential difference decreases K times).
$U_i = \frac{1}{2}CV^2$, $U_f = \frac{1}{2}C'V'^2 = \frac{1}{2}KC\left(\frac{V}{K}\right)^2 = \frac{U_i}{K}$ (energy decreases K times).
$\Delta U = U_f - U_i = \frac{1}{2}CV^2\left(\frac{1}{K} - 1\right)$
With the dielectric: $C' = KC$ and $V' = \frac{V}{K}$ (potential difference decreases K times).
$U_i = \frac{1}{2}CV^2$, $U_f = \frac{1}{2}C'V'^2 = \frac{1}{2}KC\left(\frac{V}{K}\right)^2 = \frac{U_i}{K}$ (energy decreases K times).
$\Delta U = U_f - U_i = \frac{1}{2}CV^2\left(\frac{1}{K} - 1\right)$