Three concentric spherical shells have radii a, b, and c(a < b < c) and have surface charge densities σ,-σ and σ respectively. If V_A, V_B and V_C denote the potentials of the three shells, then, for c=a+b , we have :

Correct answer: B. V_ C =V_ A ≠ V_ B aligned q_ A & =4 π a^2 σ, q_ B & =-4 π b^2 σ, q_ C & =4 π c^2 σ, c=a+b V_ A & =(1)/(4 π ∈_0)( q_ A a + q_ B b + q_ C c ) & = 2 σ a _0 V_ B & =(1)/(4 π _0)( q_ A a + q_ B b + q_ C c ) aligned aligned & =(σ)/( _0) ( a^2 b -b+c) & =(σ)/( _0) (a+ a^2 b ) V_ C & =(1)/(4 π ∈_0)( q_ A a + q_ B b + q_ C c ) & =(σ)/( _0)( a^2-b c +c)= 2 σ a _0 aligned So, V_C=V_A ≠ V_B

NEET 2009

Exam: NEET
Year: 2009
Subject: Physics
Chapter: Electrostatics

NEET 2009 · Physics · Electrostatics

Three concentric spherical shells have radii a, b, and c(a < b < c)...

Three concentric spherical shells have radii a, b, and

\mathrm{c}(\mathrm{a} < \mathrm{b} < \mathrm{c})

and have surface charge densities

\sigma,-\sigma

and

\sigma

respectively. If

V_A, V_B

and

V_C

denote the potentials of the three shells, then, for

c=a+b

, we have :

Options

A $V_C=V_B=V_A$
B $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}$
C $V_C=V_B \neq V_A$
D $\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{A}}$

Correct answer

B. $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}}$

Step-by-step solution

\begin{aligned}\mathrm{q}_{\mathrm{A}}& =4 \pi \mathrm{a}^2 \sigma,\\ \mathrm{q}_{\mathrm{B}}& =-4 \pi \mathrm{b}^2 \sigma,\\ \mathrm{q}_{\mathrm{C}}& =4 \pi \mathrm{c}^2 \sigma, \mathrm{c}=\mathbf{a}+\mathrm{b}\\ \mathrm{V}_{\mathrm{A}}& =\frac{1}{4 \pi \in_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right)\\ & =\frac{2 \sigma \mathrm{a}}{\epsilon_0}\\ \mathrm{ V}_{\mathrm{B}}& =\frac{1}{4 \pi \epsilon_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right) \end{aligned}

\begin{aligned}& =\frac{\sigma}{\epsilon_0} \left(\frac{\mathbf{a}^2}{\mathbf{b}}-\mathbf{b}+\mathrm{c}\right)\\ & =\frac{\sigma}{\epsilon_0} \left(\mathbf{a}+\frac{\mathbf{a}^2}{\mathbf{b}}\right)\\ \mathrm{V}_{\mathrm{C}}& =\frac{1}{4 \pi \in_0}\left(\frac{\mathrm{q}_{\mathrm{A}}}{\mathbf{a}}+\frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{b}}+\frac{\mathrm{q}_{\mathrm{C}}}{\mathrm{c}}\right)\\ & =\frac{\sigma}{\epsilon_0}\left(\frac{\mathrm{a}^2-\mathbf{b}}{\mathrm{c}}+\mathrm{c}\right)=\frac{2 \sigma \mathbf{a}}{\epsilon_0} \end{aligned}

So, $ V_C=V_A ≠ V_B

Practice Electrostatics on Quantrex Academy →

More from Electrostatics