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Electrostatics question

2018 · Shift 0 · Q53
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Electrostatics question

2018 · Shift 0 · Q53

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σ+ \sigma+σ, −σ- \sigma−σ and +σ+ \sigma+σ respectively. The potential of shell B is :
  1. A
    σ∈0[b2−c2c+a]{\sigma \over { \in {}_0}}\left[ {{{{b^2} - {c^2}} \over c} + a} \right]∈0​σ​[cb2−c2​+a]
  2. B
    σ∈0[a2−b2a+c]{\sigma \over { \in {}_0}}\left[ {{{{a^2} - {b^2}} \over a} + c} \right]∈0​σ​[aa2−b2​+c]
  3. C
    σ∈0[a2−b2b+c]{\sigma \over { \in {}_0}}\left[ {{{{a^2} - {b^2}} \over b} + c} \right]∈0​σ​[ba2−b2​+c]
  4. D
    σ∈0[b2−c2b+a]{\sigma \over { \in {}_0}}\left[ {{{{b^2} - {c^2}} \over b} + a} \right]∈0​σ​[bb2−c2​+a]
View written solutionFree

Correct answer: C

  1. Charges on the shells

For a spherical metal shell of radius RRR with surface charge density ρs\rho_sρs​, total charge is Q=4πR2ρs.Q=4\pi R^2 \rho_s.Q=4πR2ρs​.

So for the three concentric shells:

  • Shell AAA (radius aaa): QA=4πa2σQ_A=4\pi a^2\sigmaQA​=4πa2σ
  • Shell BBB (radius bbb): QB=−4πb2σQ_B=-4\pi b^2\sigmaQB​=−4πb2σ
  • Shell CCC (radius ccc): QC=4πc2σQ_C=4\pi c^2\sigmaQC​=4πc2σ

  1. Potential at shell BBB due to each shell

We need the potential at radius r=br=br=b.

Recall:

  • Potential due to a spherical shell at an external point r≥Rr\ge Rr≥R is V=14πε0QrV=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}V=4πε0​1​rQ​
  • Potential due to a spherical shell at an internal point r≤Rr\le Rr≤R is constant: V=14πε0QRV=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}V=4πε0​1​RQ​

Now evaluate contribution of each shell at r=br=br=b:

(i) Due to shell AAA

Since b>ab>ab>a, point is outside shell AAA:

=\frac{1}{4\pi\varepsilon_0}\frac{4\pi a^2\sigma}{b} =\frac{\sigma a^2}{\varepsilon_0 b}$$ ### (ii) Due to shell $B$ At its own surface, potential is $$V_B^{(self)}=\frac{1}{4\pi\varepsilon_0}\frac{Q_B}{b} =\frac{1}{4\pi\varepsilon_0}\frac{-4\pi b^2\sigma}{b} =-\frac{\sigma b}{\varepsilon_0}$$ ### (iii) Due to shell $C$ Since $b<c$, point is inside shell $C$, so potential is constant and equal to surface potential: $$V_C=\frac{1}{4\pi\varepsilon_0}\frac{Q_C}{c} =\frac{1}{4\pi\varepsilon_0}\frac{4\pi c^2\sigma}{c} =\frac{\sigma c}{\varepsilon_0}$$ --- 3. **Net potential of shell $B$** Add all contributions: $$V=V_A+V_B^{(self)}+V_C$$ $$V=\frac{\sigma a^2}{\varepsilon_0 b}-\frac{\sigma b}{\varepsilon_0}+\frac{\sigma c}{\varepsilon_0}$$ Factor out $\dfrac{\sigma}{\varepsilon_0}$: $$V=\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{b}-b+c\right)$$ Write $-b$ as $-\dfrac{b^2}{b}$: $$V=\frac{\sigma}{\varepsilon_0}\left(\frac{a^2-b^2}{b}+c\right)$$ --- 4. **Match with options** This matches **Option C**: $$\boxed{\frac{\sigma}{\varepsilon_0}\left[\frac{a^2-b^2}{b}+c\right]}$$ > In the options, $\in {}_0$ is clearly a typesetting error for $\varepsilon_0$. --- 5. **Comparison with stored correct answer** Stored correct answer: **C** Our derived answer: **C** So they agree.
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