Separation between the plates of a parallel plate capacitor is 5 mm . 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 mm 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 -

Question

Separation between the plates of a parallel plate capacitor is 5 mm . 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 mm 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 - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-r33tk5q8v7m3re2a.jpg" /> (A) 18.5V (B) 13.5V (C) 11.5 V (D) 6.5V

Tardigrade Admin · Accepted Answer

As the charge remain constant, hence Q=CV=C1V1 Q=(ϵ 0 A/d)V=(ϵ 0 A/d - t + (t)K)V1 ((ϵ )0 A/0.005)(25)=((ϵ )0 A/0.002 + (0.003)10)(V) V =11.5 textV

$18.5 V$

$13.5 V$

$11.5 V$

$6.5 V$

As the charge remain constant, hence
$Q = C V = C_{1} V_{1}$
$Q = \frac{ϵ _{0} A}{d} V = \frac{ϵ _{0} A}{d - t + \frac{t}{K}} V_{1}$

$\frac{( ϵ ) _{0} A}{0.005} (25) = \frac{( ϵ ) _{0} A}{0.002 + \frac{0.003}{10}} (V)$
$V = 11.5 V$

18.5V

13.5V

11.5V

6.5V

As the charge remain constant, hence Q=CV=C1​V1​ Q=dϵ0​A​V=d−t+Kt​ϵ0​A​V1​ 0.005(ϵ)0​A​(25)=0.002+100.003​(ϵ)0​A​(V) V=11.5V

$18.5 V$

$13.5 V$

$11.5 V$

$6.5 V$

As the charge remain constant, hence
$Q = C V = C_{1} V_{1}$
$Q = \frac{ϵ _{0} A}{d} V = \frac{ϵ _{0} A}{d - t + \frac{t}{K}} V_{1}$

$\frac{( ϵ ) _{0} A}{0.005} (25) = \frac{( ϵ ) _{0} A}{0.002 + \frac{0.003}{10}} (V)$
$V = 11.5 V$