A condenser is charged to a potential difference of $120 V$ . It's energy is $1 \times 10^{- 5} J$ . If battery is there and the space between plates is filled up with a dielectric medium $(ε_{r} = 5)$ . Its new energy is

$5 \times 10^{- 5} J$
$10^{- 5} J$
$3 \times 10^{- 5} J$
$2 \times 10^{- 5} J$

Question

A condenser is charged to a potential difference of $120 V$ . It's energy is $1 \times 10^{- 5} J$ . If battery is there and the space between plates is filled up with a dielectric medium $(ε_{r} = 5)$ . Its new energy is

$5 \times 10^{- 5} J$
$10^{- 5} J$
$3 \times 10^{- 5} J$
$2 \times 10^{- 5} J$

Toppr Admin · Accepted Answer

Given energy - 1-xD7-10-x2212-5-xA0-When battery is connected- V across capacitor will be constant-When dielectric is filled- -xA0-C increases by K-xA0-x2234-New-xA0-energy-xA0-12-KC-V-xA0-2-K12CV-xA0-2-Given-12CV2-1-xD7-10-x2212-5-xA0-J-K-1-xD7-10-x2212-5-xA0-J-5-xD7-1-xD7-10-x2212-5-xA0-J-5-xD7-10-x2212-5-xA0-J

A condenser is charged to a potential difference of 120V. It's energy is 1×10−5J. If battery is there and the space between plates is filled up with a dielectric medium (εr=5). Its new energy is5×10−5J10−5J3×10−5J2×10−5J

Given energy = 1×10−5 When battery is connected, V across capacitor will be constant.When dielectric is filled, C increases by K ∴New energy =12(KC)V 2=K12CV 2(Given:12CV2=1×10−5 J)=K(1×10−5) J=5×(1×10−5) J=5×10−5 J