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QJEEM18Phb820: JEE Main 2018 Physics - Class 12 JEEM18Phb820 Solution: Three concentric metal shells A, B and C of respective radii a, …

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- \sigma and +σ+ \sigma respectively. The potential of shell B is :

Multiple Choice Options

  • A σ∈0[b2−c2c+a]{\sigma \over { \in {}_0}}\left[ {{{{b^2} - {c^2}} \over c} + a} \right]
  • B σ∈0[a2−b2a+c]{\sigma \over { \in {}_0}}\left[ {{{{a^2} - {b^2}} \over a} + c} \right]
  • C σ∈0[a2−b2b+c]{\sigma \over { \in {}_0}}\left[ {{{{a^2} - {b^2}} \over b} + c} \right]
  • D σ∈0[b2−c2b+a]{\sigma \over { \in {}_0}}\left[ {{{{b^2} - {c^2}} \over b} + a} \right]

Step-by-Step Verified Solution

JEE Main 2018 (Offline) Physics - Electrostatics Question 198 English Explanation

Let charge of shell A, B and C are QA, QB and QC respectively.

Potential of B shell will be due to charge QA, QB and QC.

Here charge QA is inside of the shell B and QB is on the surface of the shell B in both cases you have to take the radius of the shell B, while calculating potential of shell B.

Charge QC is outside of the shell B so take radius of shell C for calculating potential of shell B.

∴\therefore VB = V Qa_{{Q_a}} + V Qb_{{Q_b}} + V Qc_{{Q_c}}

= 14πε0{1 \over {4\pi {\varepsilon _0}}} [4πa2(+σ)b+4πb2(−σ)b+4πc2(+σ)c]\left[ {{{4\pi {a^2}\left( { + \sigma } \right)} \over b} + {{4\pi {b^2}\left( { - \sigma } \right)} \over b} + {{4\pi {c^2}\left( { + \sigma } \right)} \over c}} \right]

= 14πε0{1 \over {4\pi {\varepsilon _0}}} ×\times 4 π\pi σ\sigma [a2b−b2b+c]\left[ {{{{a^2}} \over b} - {{{b^2}} \over b} + c} \right]

= σε0[a2−b2b+c]{\sigma \over {{\varepsilon _0}}}\left[ {{{{a^2} - {b^2}} \over b} + c} \right]

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