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Question
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
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Solution
Given: Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Formula: Electrostatic energy stored in the capacitor is given by,
`E = 1/2 CV^2`
Initial electrostatic energy:
`E_1 = 1/2 CV^2`
= `1/2 xx (600 xx 10^-12) xx (200)^2`
= `1/2 xx (600 xx 10^-12) xx 40000`
= 300 × 10−12 × 40000
= 12000 × 10−9
= 1.2 × 10−5 J
When another uncharged capacitor of 600 pF is connected:
Equivalent capacitance:
Ceq = C + C
= 600 + 600
= 1200 pF
Since both capacitors are identical, the charge is distributed equally.
Therefore, new potential difference:
`V' = 200/2`
= 100 V
New electrostatic energy:
`E_2 = 1/2 xx C_eq xx V^2`
= `1/2 xx (1200 xx 10^-12) xx (100)^2`
= `1/2 xx (1200 xx 10^-12) xx 10000`
= 600 × 10−12 × 10000
= 600 × 10−8
= 0.6 × 10−5 J
Loss in electrostatic energy = E1 − E2
= 1.2 × 10−5 − 0.6 × 10−5
= 0.6 × 10−5
= 6 × 10−6 J
Therefore, the electrostatic energy lost in the process is 6 × 10−6 J.
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