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

Question

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 $α \times 10^{- 7} J$ . The value of $α$ is ______. (combined capacitance of spheres is 200 pF). [2026]

Smart Achievers · Accepted Answer

(2)

$Q = C V = 10^{- 8} C$

$U_{1} = \frac{Q^{2}}{2 C} = \frac{10^{- 16}}{2 \times 10^{- 10}} J = 5 \times 10^{- 7} J$

$Now, when identical metal sphere is touched, charge on them becomes equal, \frac{Q}{2}$

$Energy U_{f} = \frac{{(\frac{Q}{2})}^{2}}{2 C} \times 2 = \frac{Q^{2}}{4 C} = \frac{5}{2} \times 10^{- 7} J$

$Δ U = U_{f} - U_{i} = 5 \times 10^{- 7} - \frac{5}{2} \times 10^{- 7} = 2.5 \times 10^{- 7} J$

Solution

1 $5$

2 $\frac{5}{2}$

3 $\frac{7}{2}$

4 $\frac{9}{2}$

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Solution

1 5

2 52

3 72

4 92

Ans. (2) Q=CV=10-8 C U1=Q22C=10-162×10-10 J=5×10-7 J Now, when identical metal sphere is touched, charge on them becomes equal, Q2 Energy Uf=(Q2)22C×2=Q24C=52×10-7 J ΔU=Uf-Ui=5×10-7-52×10-7=2.5×10-7 J

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1 $5$

2 $\frac{5}{2}$

3 $\frac{7}{2}$

4 $\frac{9}{2}$