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Charging of Capacitors

For a capacit...

Question

For a capacitor, the distance between two plates is $5 x$ , the electric field between them is $E_{0}$ , now dielectric slab having dielectric constant $3$ and thickness $3 x$ is placed between them in contact with one plate. In this condition what is the potential difference between its two plates?

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Solution

For a capacitor,

The distance between two places is $= 5 x$

The electric field between them $= E_{0}$

Dielectric constant $= 3$

Thickness $= 3 x$

Potential difference $= V$

$Q = C V V = \frac{Q}{C} = \frac{Q}{\frac{E_{0} E_{A}}{d}} = \frac{Q_{d}}{E_{0} E_{A}} = \frac{Q \times 5 x}{3 E_{0} \times 3 x} = \frac{5 Q}{3 E_{0} \times 3} = \frac{5 Q}{9 E_{0}}$

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