Capacitors

JEE Main level, single correct, medium

22 of 180

Question 22 of 180

A sphere of capacitance 100100 pF is charged to a potential of 100100 V. Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is α×10−7\alpha \times 10^{-7} J. The value of α\alpha is __________. (combined capacitance of spheres is 200200 pF)

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Answer: (B) 52\dfrac{5}{2}

Initial energy of the charged sphere is given by:

Ui=12CV2U_i = \dfrac{1}{2} C V^2

Ui=12×100×10−12×(100)2=5×10−7U_i = \dfrac{1}{2} \times 100 \times 10^{-12} \times (100)^2 = 5 \times 10^{-7} J

When the identical uncharged sphere is brought in contact, the charge is shared equally and the common potential becomes:

V′=V2=50V' = \dfrac{V}{2} = 50 V

Final total energy of the two spheres is:

Uf=2×(12CV′2)=CV′2U_f = 2 \times \left( \dfrac{1}{2} C V'^2 \right) = C V'^2

Uf=100×10−12×(50)2=2.5×10−7U_f = 100 \times 10^{-12} \times (50)^2 = 2.5 \times 10^{-7} J

Change in energy is the loss of energy during charge sharing:

ΔU=Ui−Uf=5×10−7−2.5×10−7=2.5×10−7\Delta U = U_i - U_f = 5 \times 10^{-7} - 2.5 \times 10^{-7} = 2.5 \times 10^{-7} J

ΔU=52×10−7\Delta U = \dfrac{5}{2} \times 10^{-7} J

Comparing with α×10−7\alpha \times 10^{-7} J, we get α=52\alpha = \dfrac{5}{2}.

Answer: 52\dfrac{5}{2}

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