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The distance between the plates of a parallel-plate capacitor is 0.05 m. A field of 3x10⁴ V/m is established between the plates. It is disconnected from the battery and an uncharged metal plate of thickness 0.01 m is inserted. What would be the potential difference (in V) if a plate of dielectric constant K=2 is introduced in place of metal plate?

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answer is 1350.

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Detailed Solution

$p.d. across capacitor V = E d = 30000 \times 0.05 = 1500 V$

$C = \frac{A \in_{0}}{d} = \frac{A \in_{0}}{0.05}$

$C' = \frac{A \in_{0}}{d - t + \frac{t}{K}} = \frac{A \in_{0}}{0.05 - 0.01 + \frac{0.01}{\infty}} = \frac{A \in_{0}}{0.04}$

$Since charge remains same$

$CV = C' V'$

$\frac{A \in_{0}}{0.05} \times 1500 = \frac{A \in_{0}}{0.04} V'$

$\Rightarrow V' = 1200 V$

$C = \frac{A \in_{0}}{d - t + \frac{t}{K}} = \frac{A \in_{0}}{0.05 - 0.01 + \frac{0.01}{2}} = \frac{A \in_{0}}{0.045}$

$Now, CV = C' V'$

$\frac{A \in_{0}}{0.05} \times 1500 = \frac{A \in_{0}}{0.045} V'$

$\Rightarrow V' = 1350 V$

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