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2. Electric Potential and Capacitance
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Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. On bringing them in contact there is
A
always a decrease in energy of the system
B
an increase in the energy of the system if $Q_1R_2 \neq Q_2R_1$
C
no change in the energy of the system
D
a decrease in energy of the system if $Q_1R_2 \neq Q_2R_1$
Solution
$\mathrm{V}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}}$
if $V_{1}=V_{2}$ then there is no loss of energy, other wise it will be.
$\therefore \frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{Q}_{2}}{\mathrm{R}_{2}} \Rightarrow \mathrm{Q}_{1} \mathrm{R}_{2}=\mathrm{Q}_{2} \mathrm{R}_{1}$ then no loss of energy.
Standard 12
Physics
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