A condenser is charged to a potential difference of 120 V- It-s energy is 1 - 10-5 J- If battery is there and the space between plates is filled up with a dielectric medium - r-5 - Its new energy is

The correct option is A 5 × 10^ 5J Given energy = 1× 10^ 5 When battery is connected, V across capacitor will be constant.When dielectric is fi ...

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Standard XII

Physics

New Capacitance in Dielectric

A condenser i...

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

10^{- 5} J

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2 \times 10^{- 5} J

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3 \times 10^{- 5} J

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5 \times 10^{- 5} J

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Solution

The correct option is A $5 \times 10^{- 5} J$
Given energy = $1 \times 10^{- 5}$
When battery is connected, $V$ across capacitor will be constant.
When dielectric is filled, $C$ increases by $K$
$∴ N e w e n e r g y = \frac{1}{2} (K C) V^{2}$
$= K \frac{1}{2} C V^{2} (G i v e n : \frac{1}{2} C V^{2} = 1 \times 10^{- 5} J)$
$= K (1 \times 10^{- 5}) J$
$= 5 \times (1 \times 10^{- 5}) J$
$= 5 \times 10^{- 5} J$

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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 is

The correct option is A 5×10−5JGiven 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

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