A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. lf oil is removed, the capacitance of capacitor becomes

Question

A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. lf oil is removed, the capacitance of capacitor becomes (A)  √ 2 C (B) 2C (C)  (C/√ 2) (D)  (C/2)

Tardigrade Admin · Accepted Answer

The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is

C = \frac{K ε _{0} A}{d} ... (i)

When dielectric (oil) is removed, so capacitance

C_{0} = \frac{ε _{0} A}{d} ... (ii)

Comparing Eqs. (i) and (ii), we get

C = K C_{0}

\Rightarrow C_{0} = \frac{C}{K} = \frac{C}{2} (∴ k = 2)

Q. A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. lf oil is removed, the capacitance of capacitor becomes

$2 C$

2C

$\frac{C}{2}$

$\frac{C}{2}$

Q. A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. lf oil is removed, the capacitance of capacitor becomes

2​C

2C

2​C​

2C​

The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is C=dKε0​A​...(i) When dielectric (oil) is removed, so capacitance C0​=dε0​A​...(ii) Comparing Eqs. (i) and (ii), we get C=KC0​ ⇒C0​=KC​=2C​(∴k=2)

$2 C$

$\frac{C}{2}$

$\frac{C}{2}$