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QJEEM19Phf8ac: JEE Main 2019 Physics - Class 12 JEEM19Phf8ac Solution: A charge Q is distributed over three concentric spherical…

A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be -

Multiple Choice Options

  • A Q(a2+b2+c2)4πε0(a3+b3+c3){{Q\left( {{a^2} + {b^2} + {c^2}} \right)} \over {4\pi {\varepsilon _0}\left( {{a^3} + {b^3} + {c^3}} \right)}}
  • B Q4πε0(a+b+c){Q \over {4\pi {\varepsilon _0}\left( {a + b + c} \right)}}
  • C Q12πε0ab+bc+caabc{Q \over {12\pi {\varepsilon _0}}}{{ab + bc + ca} \over {abc}}
  • D Q(a+b+c)4πε0(a2+b2+c2){{Q\left( {a + b + c} \right)} \over {4\pi {\varepsilon _0}\left( {{a^2} + {b^2} + {c^2}} \right)}}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 10th January Morning Slot Physics - Electrostatics Question 184 English Explanation

Potential at point P, V = kQaa+kQbb+kQcc{{k{Q_a}} \over a} + {{k{Q_b}} \over b} + {{k{Q_c}} \over c}

∵\because   Qa : Qb : Qc : : a2 : b2 : c2

{since σ\sigma a = σ\sigma b = σ\sigma c}

∴\therefore   Qa = [a2a2+b2+c2]\left[ {{{{a^2}} \over {{a^2} + {b^2} + {c^2}}}} \right] Q

Qb = [b2a2+b2+c2]\left[ {{{{b^2}} \over {{a^2} + {b^2} + {c^2}}}} \right] Q

Qc = [c2a2+b2+c2]\left[ {{{{c^2}} \over {{a^2} + {b^2} + {c^2}}}} \right] Q

V = Q4π∈0[(a+b+c)a2+b2+c2]{Q \over {4\pi { \in _0}}}\left[ {{{\left( {a + b + c} \right)} \over {{a^2} + {b^2} + {c^2}}}} \right]

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