Question

The distance between the plates of a parallel plate condenser is $8mm$ and potential difference $120$ volts. If a $6mm$ 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 potential difference across the condenser will be 320 volts
• D. the potential difference across the condenser will be 45 volts

Answer 1

Answer: (D)

If nothing is said, it is considered that battery is disconnected.
Hence, charge remain the same.
Also, $V_{\text{air }}=\frac{\sigma }{{\epsilon }_0}\times d$
and $V_{\text{medium }}=\frac{\sigma }{{\epsilon }_0}\left(d-t+\frac{t}{k}\right)$
$\Rightarrow \frac{V_{\text{medium }}}{V_{\text{air }}}=\frac{\left(d-t+\frac{t}{k}\right)}{d}$
$\Rightarrow \frac{V_{\text{medium }}}{120}=\frac{\left(8-6+\frac{6}{6}\right)}{8}$
$\Rightarrow V_{di}=45V$