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Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities \sigma, - \sigma and \sigma respectively. If VA, VB and Vc denotes the potentials of the three shells, then for c = a + b, we have

  • Option 1)

    V_{C}=V_{B}\neq V_{A}

  • Option 2)

    V_{C}= V_{B}\neq V_{A}

  • Option 3)

    V_{C}=V_{B}=V_{A}

  • Option 4)

    V_{C}=V_{A}\neq V_{B}

 

Answers (1)

As we learnt in

Outside the sphere (P lies outside the sphere) -

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

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

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

-

 

 

V_{A} = \frac{kq_{A}}{a}+\frac{kq_{B}}{b}+\frac{kq_{C}}{c}

= \frac{k.\sigma .4\pi a^{2}}{a}-\frac{k.\sigma .4\pi b^{2}}{b}+\frac{k.\sigma .4\pi c^{2}}{c}

V_{A}=k.\sigma .4\pi \left ( a-b+c \right )

V_{B}= \frac{kq_{A}}{b}+\frac{kq_{B}}{b}+\frac{kq_{C}}{c}

=\frac{k.\sigma .4\pi a^{2}}{b}+\frac{k\left ( -\sigma .4\pi b^{2} \right )}{b}+\frac{k.\sigma .4\pi c^{2}}{c}

= 4\pi k\sigma \left [ \frac{a^{2}}{b}-b+c\right ]

= 4\pi k\sigma \left ( \frac{a^{2}-b^{2}+bc}{b} \right )

V_{c} =\frac{k.q_{A}}{a}+\frac{k.q_{B}}{b}+\frac{k.q_{C}}{c}

= \frac{k.\sigma 4\pi a^{2}}{c}-\frac{k.\sigma 4\pi b^{2}}{c}+\frac{k.\sigma 4\pi c^{2}}{c}

= 4\pi k\sigma \frac{\left [ a^{2}-b^{2}+c^{2}\right ]}{c}

4\pi k\sigma \left [ \frac{\left ( a-b \right ).\left ( a+b \right )}{c} +c\right ]

=4\pi k\sigma \left [ a-b+c \right ]\left [ \therefore a+b=c \right ]

\therefore V_{A}=V_{C}\neq V_{B}


Option 1)

V_{C}=V_{B}\neq V_{A}

Incorrect

Option 2)

V_{C}= V_{B}\neq V_{A}

Incorrect

Option 3)

V_{C}=V_{B}=V_{A}

Incorrect

Option 4)

V_{C}=V_{A}\neq V_{B}

Correct

Posted by

subam

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