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  ______.

Options

• 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:

V_i = `1/2` (10C)V²

Here, V = constant and 'C' becomes 1.5 times

So, U_f = `1/2"C"_"f""V"_"f"^2`

= `1/2xx15"C"xx"V"^2`

= `15xx1/2"CV"^2xx10/10`

= `1.5xx1/2(10"C")"V"^2`

= 1.5 V_i

`(Delta"V")/"V"xx100`

= `(0.5"V"_"i")/"V"_"i"xx100`

= 50%