The electrostatic potential ($\phi_{\mathrm{r}}$) of a spherical symmetric system, kept at origin, is shown in the adjacent figure, and given as $\phi_{\mathrm{r}}=\frac{\mathrm{q}}{4 \pi \epsilon_{0} \mathrm{r}} \quad\left(\mathrm{r} \geq \mathrm{R}_{0}\right)$ $\phi_{\mathrm{r}}=\frac{\mathrm{q}}{4 \pi \epsilon_{0} \mathrm{R}_{0}} \quad\left(\mathrm{r} \leq \mathrm{R}_{0}\right)$ Which of the following option(s) is/are correct?

(A) For spherical region $\mathrm{r} \leq \mathrm{R}_{0}$, total electrostatic energy stored is zero.; (B) Within $\mathrm{r}=2 \mathrm{R}_{0}$, total charge is q.; (C) There will be no charge anywhere except at $\mathrm{r}=\mathrm{R}_{0}$.; (D) Electric field is discontinuous at $\mathrm{r}=\mathrm{R}_{0}$.

JEE Advanced · 2006 · 09 Apr 2006 · Electrostatics

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The electrostatic potential (

ϕ_{r}

) of a spherical symmetric system, kept at origin, is shown in the adjacent figure, and given as

ϕ_{r} = \frac{q}{4 π ϵ _{0} r} (r \geq R_{0})

ϕ_{r} = \frac{q}{4 π ϵ _{0} R _{0}} (r \leq R_{0})

Which of the following option(s) is/are correct?

(A)For spherical region $r \leq R_{0}$ , total electrostatic energy stored is zero.✓ Correct
(B)Within $r = 2 R_{0}$ , total charge is q.✓ Correct
(C)There will be no charge anywhere except at $r = R_{0}$ .✓ Correct
(D)Electric field is discontinuous at $r = R_{0}$ .✓ Correct

Answer: (A), (B), (C), (D)

Solution

The given potential distribution is characteristic of a charged spherical conductor of radius

R_{0}

with total charge q. For

r \leq R_{0}

, the potential is constant, which means the electric field

E = - \frac{d ϕ}{d r} = 0

inside. Since there is no electric field inside, the electrostatic energy stored is zero (option A correct). All charge q resides on the surface at

r = R_{0}

, so within any radius

r \geq R_{0}

(including

r = 2 R_{0}

), the total enclosed charge is q (option B correct). Since it's a conductor, all charge is on the surface at

r = R_{0}

with none elsewhere (option C correct). At the surface, the electric field jumps from 0 (inside) to

\frac{q}{4 π ϵ _{0} R _{0}^{2}}

(outside), making it discontinuous (option D correct).

JEE Advanced · 2006 · 09 Apr 2006 · One or more correct options

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JEE Advanced · 2006 · 09 Apr 2006 · Electrostatics

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