JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 18)
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :
508 pJ
692 pJ
560 pJ
600 pJ
Explanation
Initial energy of capacitor
Ui = $${1 \over 2}$$ $${{{v^2}} \over c}$$
= $${1 \over 2}$$ $$ \times $$ $${{120 \times 120} \over {12}}$$ = 600 J
Since battery is disconnected so charge remain same.
Final energy of capacitor
Uf = $${1 \over 2}{{{v^2}} \over c}$$
= $${1 \over 2} \times {{120 \times 120} \over {12 \times 6.5}}$$ = 92
W + Uf = Ui
W = 508 J
Ui = $${1 \over 2}$$ $${{{v^2}} \over c}$$
= $${1 \over 2}$$ $$ \times $$ $${{120 \times 120} \over {12}}$$ = 600 J
Since battery is disconnected so charge remain same.
Final energy of capacitor
Uf = $${1 \over 2}{{{v^2}} \over c}$$
= $${1 \over 2} \times {{120 \times 120} \over {12 \times 6.5}}$$ = 92
W + Uf = Ui
W = 508 J
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