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Three isolated metal spheres A, B, C have radius R, 2R, 3R respectively, and the same charge, $Q . U_A, U_B$ and $U_C$ be the energy density just outside the surface of the spheres. The relation between $\mathrm{U}_{\mathrm{A},} \mathrm{U}_{\mathrm{B}}$ and $\mathrm{U}_{\mathrm{C}}$ is
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The correct answer is:
$\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}$
The correct option is (B).
Concept: Energy density in the free space is given by, $\mathrm{U}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2$
The electric field for an isolated sphere is $E=\frac{Q}{2 \pi \varepsilon_0 r^2}$
Therefore, Energy density $U \propto \frac{1}{\mathrm{r}^4}$
And, $\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}$ is the correct trend.
Concept: Energy density in the free space is given by, $\mathrm{U}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2$
The electric field for an isolated sphere is $E=\frac{Q}{2 \pi \varepsilon_0 r^2}$
Therefore, Energy density $U \propto \frac{1}{\mathrm{r}^4}$
And, $\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}$ is the correct trend.
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