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: A. V_ C =V_ A ≠ V_ B Here, aligned & V_ A =(1)/(4 π _0) σ 4 π a^2 a -(1)/(4 π _0) σ 4 π b^2 ~b & +(1)/(4 π _0) · (σ 4 π c^2)/(c) & =(σ)/( _0)(a-b+c)=(σ)/( _0)(2 a) ( c=a+b) & aligned aligned & V_ B =(1)/(4 π _0) · σ 4 π a^2 a -(1)/(4 π _0) σ 4 π b^2 ~b &+(1)/(4 π _0) · σ 4 π c^2 c aligned aligned & =(σ)/( _0)((a^2)/(c)-b+c)=(σ)/( _0)(2 a)( c=a+b) & and V_C=(1)/(4 π _0) · (σ 4 π a^2)/(c)-(1)/(4 π _0) (σ 4 π b^2)/(c) & +(1)/(4 π _0) · (σ 4 π c^2)/(c) & =(σ)/( _0)((a^2)/(c)-(b^2)/(c)+c)=(σ)/( _0)(2 a)( c=a+b) & aligned Hence, V_ A =V_ C ≠ 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

c(a < b < 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

Three concentric spherical shells have radii a, b and c(a < b < c) and have surface charge densities σ,-σ and σ

Options

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

Correct answer

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

Step-by-step solution

Here,

\begin{aligned}& \mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}}\\ & +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c}\\ & =\frac{\sigma}{\varepsilon_0}(a-b+c)=\frac{\sigma}{\varepsilon_0}(2 a) \quad(\because c=a+b) \\ & \end{aligned}

\\

\begin{aligned}& \mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}}\\ &+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{c}^2}{\mathrm{c}} \end{aligned}

$\begin{aligned}& =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-b+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\ & \text{\text{ and } }V_C=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi a^2}{c}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi b^2}{c}\\ & +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c}\\ & =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-\frac{b^2}{c}+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\ & \end{aligned}

Hence,

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

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