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

  • Option 1)

    \frac{\sigma }{\varepsilon _{0}}\left ( a-b-c \right ),\frac{\sigma }{\varepsilon _{0}}\left ( \frac{a^{2}}{b}-b+c \right )

     

     

     

     

     

     

  • Option 2)

    \left ( a-b-c \right ),\frac{a^{2}}{c}

  • Option 3)

    \frac{\varepsilon _{0} }{\sigma }\left ( a-b-c \right ),\frac{\varepsilon _{0} }{\sigma }\left ( \frac{a^{2}}{c}-b+c \right )

  • Option 4)

    \frac{\sigma }{\varepsilon _{0}}\left ( \frac{a^{2}}{c}-\frac{b^{2}}{c}+c \right ),\frac{\sigma }{\varepsilon _{0}}\left ( a-b+c \right )

 

Answers (1)

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As we learned

 

Potential Due to 3 Concentric Spheres -

The figure shows three conducting concentric shell of radii a, b and c (a < b < c) having charges Qa, Qb and Qc respectively

- wherein

 

Potential at A;

V_{A}= \frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q_{a}}{a}+\frac{Q_{b}}{b} +\frac{Q_{c}}{c}\right ]

Potential at B;

V_{B}= \frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q_{a}}{b}+\frac{Q_{b}}{b} +\frac{Q_{c}}{c}\right ]

Potential at C;

V_{C}= \frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q_{a}}{c}+\frac{Q_{b}}{c} +\frac{Q_{c}}{c}\right ]

 

 Suppose charges on A, B and C are qa,qb and qc

Respectively, so \sigma _{A}=\sigma =\frac{q_{a}}{4\pi a^{2}}\Rightarrow q_{a}=\sigma \times 4\pi a^{2},

\sigma _{B}=-\sigma =\frac{q_{b}}{4\pi b^{2}}\Rightarrow q_{b}=-\sigma \times 4\pi b^{2},                                                                              

and  \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     

V_{A}=\left ( V_{A} \right )_{Surface}+\left ( V_{B} \right )_{in}+\left ( V_{C} \right )_{in}=\frac{1}{4\pi \varepsilon _{0}}\left [ \frac{q_{a}}{a}+\frac{q_{b}}{b} +\frac{q_{c}}{c}\right ]=\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 ]

V_{A}=\frac{\sigma }{\varepsilon _{0}}\left [ a-b-c \right ]                                

Potential at the surface of B

V_{B}=\left ( V_{A} \right )_{Out}+\left ( V_{B} \right )_{Surface}+\left ( V_{C} \right )_{in}=\frac{1}{4\pi \varepsilon _{0}}\left [ \frac{q_{a}}{a}+\frac{q_{b}}{b} +\frac{q_{c}}{c}\right ]=\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 ]

=\frac{\sigma }{\varepsilon _{0}}\left [\frac{a^{2}}{b}-b+c \right ]


Option 1)

\frac{\sigma }{\varepsilon _{0}}\left ( a-b-c \right ),\frac{\sigma }{\varepsilon _{0}}\left ( \frac{a^{2}}{b}-b+c \right )

 

 

 

 

 

 

Option 2)

\left ( a-b-c \right ),\frac{a^{2}}{c}

Option 3)

\frac{\varepsilon _{0} }{\sigma }\left ( a-b-c \right ),\frac{\varepsilon _{0} }{\sigma }\left ( \frac{a^{2}}{c}-b+c \right )

Option 4)

\frac{\sigma }{\varepsilon _{0}}\left ( \frac{a^{2}}{c}-\frac{b^{2}}{c}+c \right ),\frac{\sigma }{\varepsilon _{0}}\left ( a-b+c \right )

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SudhirSol

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