Electrostatic Potential and Capacitance - Result Question 27
29. 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?
[2015]
(a) The energy stored in the capacitor decreases $K$ times.
(b) The chance in energy stored is
$ \frac{1}{2} CV^{2}(\frac{1}{K}-1) $
(c) The charge on the capacitor is not conserved.
(d) The potential difference between the plates decreases $K$ times.
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Answer:
Correct Answer: 29. (c)
Solution:
- (c) Capacitance of the capacitor, $C=\frac{Q}{V}$
After inserting the dielectric, new capacitance $C^{\prime}=K . C$
New potential difference
$V^{\prime}=\frac{V}{K}$
$U_i=\frac{1}{2} CV^{2}=\frac{Q^{2}}{2 C} \quad(\because Q=CV)$
$U_f=\frac{Q^{2}}{2 C}=\frac{Q^{2}}{2 KC}=\frac{C^{2} V^{2}}{2 KC}=(\frac{U_i}{K})$
$\Delta U=U_f-U_i=\frac{1}{2} CV^{2}{\frac{1}{K}-1}$
As the capacitor is isolated, so charge will remain conserved and p.d. between two plates of the capacitor
$ V^{\prime}=\frac{Q}{KC}=\frac{V}{K} $
When the battery remains connected, the capacity of capacitor and charge stored is changed. When the battery disconnected, charge remains same but potential and capacitance are changed.
