First slide

Capacitance

Question

In the arrangement shown, when a dielectric slab having dielectric constant K is inserted in the capaciter of capacitance C, the reading of the volt meter becomes 1/5th of previous value.Then value of K is

Moderate

A

3
B

4
C

2
D

3.5

Solution

Initial reading, $V = \frac{\frac{2 C}{3} . E}{C} = \frac{2 E}{3}$
Q¹ = final charge on the capacitor = $\frac{CK \times 2 C}{(CK + 2 C)} \times E = \frac{2 KCE}{(K + 2)}$
$∴$ V' =Final potential = $\frac{Q^{1}}{CK} = \frac{2 E}{K + 2}$
Since V' = ½ V we can write $\frac{2 E}{K + 2} = \frac{1}{2} \times \frac{2 E}{3} \Rightarrow K = 4$

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