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Physics · Capacitors and R-C Circuits

JEE Main 2026 — 6 April, Evening Shift — Question 12

A sphere of capacitance 100 pF is charged to a potential of 100 V. Another identical undercharged 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 200 pF).

  1. Option A:

    5

  2. Option B:

    2.5

    Correct
  3. Option C:

    46060

  4. Option D:

    3

Answer: B

Step-by-step solution

Initial charge Q=CV=100×10−12×100=10−8Q = CV = 100\times10^{-12}\times100 = 10^{-8} C. Initial energy Ui=12CV2=0.5×100×10−12×104=5×10−7U_i = \frac12 CV^2 = 0.5\times100\times10^{-12}\times10^4 = 5\times10^{-7} J. After contact, potential Vf=Q/Ctotal=10−8/(200×10−12)=50V_f = Q/C_{total} = 10^{-8}/(200\times10^{-12}) = 50 V. Final energy Uf=12CtotalVf2=0.5×200×10−12×2500=2.5×10−7U_f = \frac12 C_{total} V_f^2 = 0.5\times200\times10^{-12}\times2500 = 2.5\times10^{-7} J. Change magnitude = 2.5×10−72.5\times10^{-7} J, so α=2.5\alpha = 2.5.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Physics
Chapter
Capacitors and R-C Circuits
Topic
Capacitance and Different types of Capacitors