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Question: Three concentric metallic spheres A, B and C have radii a, b and c\((a < b < c)\) and surface charge...

Three concentric metallic spheres A, B and C have radii a, b and c(a<b<c)(a < b < c) and surface charge densities on them are σ,−σ\sigma, - \sigma and σ\sigma respectively. The valves of VAV_{A} and VBV_{B} will be

A

σε0(a−b−c),σε0(a2b−b+c)\frac{\sigma}{\varepsilon_{0}}(a - b - c),\frac{\sigma}{\varepsilon_{0}}\left( \frac{a^{2}}{b} - b + c \right)

B

(a−b−c),a2c(a - b - c),\frac{a^{2}}{c}

C

ε0σ(a−b−c),ε0σ(a2c−b+c)\frac{\varepsilon_{0}}{\sigma}(a - b - c),\frac{\varepsilon_{0}}{\sigma}\left( \frac{a^{2}}{c} - b + c \right)

D

σε0(a2c−b2c+c)\frac{\sigma}{\varepsilon_{0}}\left( \frac{a^{2}}{c} - \frac{b^{2}}{c} + c \right)andσε0(a−b+c)\frac{\sigma}{\varepsilon_{0}}(a - b + c)

Answer

σε0(a−b−c),σε0(a2b−b+c)\frac{\sigma}{\varepsilon_{0}}(a - b - c),\frac{\sigma}{\varepsilon_{0}}\left( \frac{a^{2}}{b} - b + c \right)

Explanation

Solution

Suppose charges on A, B and C are qa,qbq_{a},q_{b} and qcq_{c}

Respectively, so σA=σ=qa4πa2⇒qa=σ×4πa2\sigma_{A} = \sigma = \frac{q_{a}}{4\pi a^{2}} \Rightarrow q_{a} = \sigma \times 4\pi a^{2},

σB=−σ=qb4πb2⇒qb=−σ×4πb2\sigma_{B} = - \sigma = \frac{q_{b}}{4\pi b^{2}} \Rightarrow q_{b} = - \sigma \times 4\pi b^{2}

and σC=σ=qc4πc2⇒qc=σ×4πc2\sigma_{C} = \sigma = \frac{q_{c}}{4\pi c^{2}} \Rightarrow q_{c} = \sigma \times 4\pi c^{2}

Potential at the surface of A

VA=(VA)surface+(VB)in+(VC)in=14πε0[qaa+qbb+qcc]V_{A} = (V_{A})_{\text{surface}} + (V_{B})_{\text{in}} + (V_{C})_{\text{in}} = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{q_{a}}{a} + \frac{q_{b}}{b} + \frac{q_{c}}{c} \right\rbrack

=14πε0[σ×4πa2a+(−σ)×4πb2b+σ×4πc2c]= \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \left[ \frac { \sigma \times 4 \pi a ^ { 2 } } { a } + \frac { ( - \sigma ) \times 4 \pi b ^ { 2 } } { b } + \frac { \sigma \times 4 \pi c ^ { 2 } } { c } \right] VA=σε0[a−b−c]]V_{A} = \frac{\sigma}{\varepsilon_{0}}\left\lbrack a - b - c\rbrack \right\rbrack

Potential at the surface of B

VB=(VA)out+(VB)surface+(VC)in=14πε0[qab+qbb+qcc]=14πε0[σ×4πa2b−σ×4πb2b+σ×4πc2c]=σε0[a2b−b+c]V_{B} = (V_{A})_{\text{out}} + (V_{B})_{\text{surface}} + (V_{C})_{\text{in}} = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{q_{a}}{b} + \frac{q_{b}}{b} + \frac{q_{c}}{c} \right\rbrack = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{\sigma \times 4\pi a^{2}}{b} - \frac{\sigma \times 4\pi b^{2}}{b} + \frac{\sigma \times 4\pi c^{2}}{c} \right\rbrack = \frac{\sigma}{\varepsilon_{0}}\left\lbrack \frac{a^{2}}{b} - b + c \right\rbrack