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Physics Question 40 – JEE-MAIN 2026

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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. The value of α is _______ (combined capacitance of spheres is 200 pF)

Recall the formula for energy stored in a capacitor and the principle of charge conservation when conductors are connected.

Step 1: Calculate Initial Energy✦ Active

The initial capacitance of the first sphere is C1=100 pF=100×10−12 F=10−10 F. It is charged to a potential V1=100 V. The initial charge on this sphere is Q1=C1V1=(10−10 F)(100 V)=10−8 C. The initial energy stored in the system (only the first sphere) is:

Uinitial=12C1V12=12(10−10 F)(100 V)2=12×10−10×104=12×10−6 J=5×10−7 J
Step 2: Calculate Final Energy○ Expand

When the identical uncharged sphere (C2=100 pF) is brought into contact with the charged sphere, the total capacitance of the combined system is Ctotal=C1+C2=100 pF+100 pF=200 pF=2×10−10 F. The total charge in the system remains conserved, Qtotal=Q1=10−8 C. The final common potential Vfinal is:

Vfinal=QtotalCtotal=10−8 C2×10−10 F=50 V

The final total energy stored in the combined system is:

Ufinal=12CtotalVfinal2=12(2×10−10 F)(50 V)2=10−10×2500=2.5×10−7 J
Step 3: Determine Change in Energy and α○ Expand

The change in total energy stored is ΔU=Ufinal−Uinitial:

ΔU=2.5×10−7 J−5×10−7 J=−2.5×10−7 J

Since energy is lost during charge redistribution, and the options are positive, the question asks for the magnitude of the energy lost. The magnitude of the change in energy is |ΔU|=2.5×10−7 J. Comparing this with α×10−7 J, we find the value of α:

α=2.5=52
💡 Teacher's Secret Hint

Remember that energy is always lost when charge is redistributed between conductors at different potentials. The 'change' in energy often refers to the magnitude of this loss in such contexts, especially when options are positive.

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