Capacitors

JEE Main level, single correct, medium

139 of 180

Question 139 of 180

A capacitor with capacitance 5μF5\mu F is charged to 5μC5\mu C . If the plates are pulled apart to reduce the capacitance to 2μF2\mu F, how much work is done?

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Answer: (D) 3.75×10–6J3.75\times {10}^{–6}J

Uinitial=Q22C1{U}_{initial}=\dfrac{{Q}^{2}}{2{C}_{1}}
Ufinal=Q22C2{U}_{final}=\dfrac{{Q}^{2}}{2{C}_{2}}
Work done =Ufinal−Uinitial={U}_{final}-{U}_{initial}
=Q22C2−Q22C1=\dfrac{{Q}^{2}}{2{C}_{2}}-\dfrac{{Q}^{2}}{2{C}_{1}}
=(5×10−6)22×2×10−6−(5×10−6)22×5×10−6=\dfrac{{\left(5\times {10}^{-6}\right)}^{2}}{2\times 2\times {10}^{-6}}-\dfrac{{\left(5\times {10}^{-6}\right)}^{2}}{2\times 5\times {10}^{-6}}
=(254−52)×10−6=\left(\dfrac{25}{4}-\dfrac{5}{2}\right)\times {10}^{-6}
=154×10−6=3.75×10−6J=\dfrac{15}{4}\times {10}^{-6}=3.75\times {10}^{-6}J

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