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  • Solve it, Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σ, −σ and +σ respectively.The potential of shell B is :

Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities +σ, −σ and +σ respectively.The potential of shell B is :

  • Option 1)

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

  • Option 2)

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

  • Option 3)

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

  • Option 4)

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

 

Answers (2)

best_answer

As we learnt that

 

Outside the sphere (P lies outside the sphere) -

\dpi{100} E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}=\frac{\sigma R^{2}}{\epsilon _{0}r^{2}}

V_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r}=\frac{\sigma R^{2}}{\epsilon _{0}r}

 

- wherein

\sigma - surface charge density.

 

 

At the surface of Sphere -

V=R

E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}=\frac{\sigma }{\epsilon _{0}}

V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{\sigma R}{\epsilon _{0}}

-

 

 

V_{B}=\frac{1}{4\pi \epsilon _{o}}[\frac{4\pi a^{2}\sigma }{b}-\frac{4\pi b^{2}\sigma }{b}+\frac{4\pi c^{2}\sigma }{b}]

V_{B}=\frac{\sigma }{c_{o}}(\frac{a^{2}-b^{2}}{b}+c)

 


Option 1)

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

This is incorrect

Option 2)

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

This is incorrect

Option 3)

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

This is correct

Option 4)

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

This is incorrect

Posted by

Aadil

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