Physics Circle home Student Login

Home›Chapters›Capacitors PYQs›Charge Sharing & Loss of Energy›Question 5

JEE Main 2026 (6 Apr, Shift 2) · Charge Sharing & Loss of Energy

Charge Sharing & Loss of Energy question from JEE Main 2026 (6 Apr, Shift 2)

Q5JEE Main 2026 · 6 APR · SHIFT 2Easy

Charge Sharing & Loss of Energy

A sphere of capacitance 100 pF is charged to a potential of 100 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 J\alpha \times 10^{-7}\text{ J}. The value of α\alpha is ______. (combined capacitance of spheres is 200 pF)

Show answer and solution

Answer: (b)

  1. Ui=12CV2=12×100×10−12×1002=5×10−7U_i = \frac{1}{2}CV^2 = \frac{1}{2} \times 100 \times 10^{-12} \times 100^2 = 5 \times 10^{-7} J
  2. Same charge on 200 pF: Uf=Q22(2C)=Ui2=2.5×10−7U_f = \frac{Q^2}{2(2C)} = \frac{U_i}{2} = 2.5 \times 10^{-7} J
  3. Change =2.5×10−7= 2.5 \times 10^{-7} J, so α=52\alpha = \frac{5}{2}, option (b).
Report a mistake

More questions on Charge Sharing & Loss of Energy