Separation between the plates of a parallel plate capacitor is 5 m m . This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery

Question

Separation between the plates of a parallel plate capacitor is $5 m m$ . This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference $25 V$ using a battery. The battery is then disconnected and a dielectric slab of thickness $3 m m$ and dielectric constant $K = 10$ is placed between the plates as shown. Potential difference between the plates after the dielectric slab has been introduced is -

Options:

Answer:

C

Solution:

As the charge remain constant, hence

$Q = C V = C_{1} V_{1}$

$Q = \frac{ε_{0} A}{d} V = \frac{ε_{0} A}{d - t + \frac{t}{K}} V_{1}$

$\frac{ε_{0} A}{0.005} (25) = \frac{ε_{0} A}{0.002 + \frac{0.003}{10}} (V)$

$V = 11.5 V$

TestHub · Accepted Answer

As the charge remain constant, hence Q = C V = C 1 V 1 Q = ε 0 A d V = ε 0 A d - t + t K V 1 ε 0 A 0.005 25 = ε 0 A 0.002 + 0.003 10 V V ⁡ = 11.5 V

Physics - Electrostatics Question with Solution | TestHub

Options:

Doubts & Discussion