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 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
- A
The charge on the condenser will be doubled
- B
The charge on the condenser will be reduced to half
- C
The $P.D.$ across the condenser will be $320\;volts$
- D
The $P.D.$ across the condenser will be $45\;volts$
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$K(x) = K_0 + \lambda x$ ( $\lambda =$ constant)
The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation
The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
$K(x) = K_0 + \lambda x$ ( $\lambda =$ constant)
The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation
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