Question

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will ______.

पर्याय

• increase by 50%

• decrease by 15%

• increase by 25%

• increase by 33%

Answer 1

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will increase by 50%.

Explanation:

Potential energy stored in air capacitor of capacitance C, charged to potential difference V is given by:

U = `1/2"CV"^2`

If the above capacitor is filled completely with a dielectric of dielectric constant K, keeping the potential difference same, then potential energy of capacitor becomes

U' = `1/2`KCV² = KU

Given:

Initially K of dielectric = 10

U_i ⇒ Initial potential energy = `1/2` × 10 × CV²

U_f ⇒ Final potential energy = `1/2` × 15 × CV² 

percentage increase in energy of capacitor

= `(U_f - U_i)/U_i` × 100

⇒ percentage increase

= `(1/2"CV"^2[15-10])/(1/2"CV"^2xx10) xx10`

= `5/10xx100`

= 50%