JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 4)
A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and
is conneced 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) _____
Answer
6
Explanation
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Ui = $${1 \over 2}CV_0^2$$
Uf = $${1 \over 2} \times 2C \times {\left( {{{{V_0}} \over 2}} \right)^2}$$
$$\Delta $$E = $${1 \over 2}CV_0^2$$ - $${1 \over 2} \times 2C \times {\left( {{{{V_0}} \over 2}} \right)^2}$$
= $${{{CV_0^2} \over 4}}$$
= $${1 \over 4} \times 60 \times {10^{ - 12}} \times 4 \times {10^2}$$
= 6 $$ \times $$ 10-9 = 6 nJ
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