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The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then
The charge on the condenser will be doubled
The charge on the condenser will be reduced to half
The $P.D.$ across the condenser will be $320\;volts$
The $P.D.$ across the condenser will be $45\;volts$
Solution
(d)If nothing is said, it is considered that battery is disconnected. Hence charge remain the same
Also ${V_{air}} = \frac{\sigma }{{{\varepsilon _0}}} \times d$ and ${V_{medium}} = \frac{\sigma }{{{\varepsilon _0}}}(d – t + \frac{t}{k})$
$==>$ $\frac{{{V_m}}}{{{V_a}}} = \frac{{(d – t + \frac{t}{k})}}{d}$ $==>$ $\frac{{{V_m}}}{{120}} = \frac{{(8 – 6 + \frac{6}{6})}}{8} $ $==>$ $ V_m = 45\,V$
