Two concentric spherical shells of radius R 1 and R 2 (R 2 > R 1 ) are having uniformly distributed charges Q 1 and Q 2 respectively. Find out total energy of the system.
Text Solution
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- Step 1: The potential due to a uniformly charged spherical shell at a point outside the shell (radius > R) is given by:
- Step 2: For the inner shell with charge Q1 at radius R1, the potential at the radius R2 (where the outer shell is) is:
- Step 3: The energy U1 of the charge Q2 in the potential V1 is given by:
- Step 4: Now, we need to calculate the energy U2 of Q1 in the potential due to Q2 at radius R1:
- Step 5: The energy U2 is:
- Step 6: Adding these two energies gives the total energy of the system:
- Final Answer: Thus, the total energy of the system is:
V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}
V_{1}(R_{2}) = \frac{1}{4\pi\epsilon_0} \frac{Q_{1}}{R_{2}}
U_{1} = Q_{2} V_{1}(R_{2}) = Q_{2} \left( \frac{1}{4\pi\epsilon_0} \frac{Q_{1}}{R_{2}} \right)
V_{2}(R_{1}) = \frac{1}{4\pi\epsilon_0} \frac{Q_{2}}{R_{1}}
U_{2} = Q_{1} V_{2}(R_{1}) = Q_{1} \left( \frac{1}{4\pi\epsilon_0} \frac{Q_{2}}{R_{1}} \right)
U_{total} = U_{1} + U_{2} = Q_{2} \left( \frac{1}{4\pi\epsilon_0} \frac{Q_{1}}{R_{2}} \right) + Q_{1} \left( \frac{1}{4\pi\epsilon_0} \frac{Q_{2}}{R_{1}} \right)
U_{total} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q_{1}Q_{2}}{R_{2}} + \frac{Q_{2}Q_{1}}{R_{1}} \right)
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