JEE PYQ: Capacitance Question 21
Question 21 - 2020 (07 Jan Shift 2)
A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)
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Answer: 6
Solution
$U_i = \frac{1}{2}CV_0^2 = \frac{1}{2} \times 60 \times 10^{-12} \times 400 = 12 \times 10^{-9}$ J. $U_f = \frac{1}{2} \times 2C \times \left(\frac{V_0}{2}\right)^2 = \frac{1}{4} \times 60 \times 10^{-12} \times (20)^2 = 6 \times 10^{-9}$ J. Energy lost $= 12 - 6 = 6$ nJ.
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