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 $\alpha \times 10^{- 7}\text{ }J$. The value of $\alpha$ is $\_\_\_\_$ . (combined capacitance of spheres is 200 pF ).
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Solution
Sol. [Key Concept: The two identical spheres share the charge equally on contact; the energy changes from $\frac{1}{2}\ {CV}^{2}$ to $2 \times \frac{1}{2}\ C(\frac{V}{2})^{2}$, so find $\Delta\ E$.]
$${Q_{1} = CV = 100 \times 100pC = 10^{4}pC }{V_{1} = V_{2} = \frac{V}{2} = 50 }{\varepsilon_{i} = \frac{1}{2} \times 100 \times (100)^{2}pJ = \varepsilon }{\varepsilon_{f} = 2 \times \frac{1}{2} \times 100 \times (50)^{2} = \frac{\varepsilon}{2} }{\Delta\varepsilon = \frac{\varepsilon}{2} = 100 \times (50)^{2}pJ }{= 25 \times 10^{4} \times 10^{- 12} = 2.5 \times 10^{- 7} }$$
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