MHT CET · Physics · Electrostatics
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
- A \(\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}<\mathrm{U}_{\mathrm{C}}\)
- B \(\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}\)
- C \(\mathrm{U}_{\mathrm{A}}<\mathrm{U}_{\mathrm{B}}<\mathrm{U}_{\mathrm{C}}\)
- D \(\mathrm{U}_{\mathrm{A}}<\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{U}_{\mathrm{A}}>\mathrm{U}_{\mathrm{B}}>\mathrm{U}_{\mathrm{C}}\)
Step-by-step Solution
Detailed explanation
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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