In the figure, the inner (shaded) region represents a sphere of radius , within which the electrostatic charge density varies with the radial distance from the center as where is positive. In the spherical shell of outer radius , the electrostatic c

Question

In the figure, the inner (shaded) region $A$ represents a sphere of radius $r_{A} = 1$ , within which the electrostatic charge density varies with the radial distance $r$ from the center as $ρ_{A} = k r,$ where $k$ is positive.

In the spherical shell $B$ of outer radius $r_{B}$ , the electrostatic charge density varies as $ρ_{B} = \frac{2 k}{r} .$

Assume that dimensions are taken care of. All physical quantities are in their SI units.

Which of the following statement(s) is (are) correct? [2022]

Smart Achievers · Accepted Answer

(2)

We have

$q_{A} = \int_{0}^{1} k r (4 π r^{2} d r) = 4 π k \int_{0}^{1} r^{3} d r = \frac{4 π k}{4} {[r^{4}]}_{0}^{1} = π k$

$q_{B} = \int_{1}^{r} \frac{2 k}{r} 4 π r^{2} d r = 8 π k \int_{1}^{r} r d r = 8 π k (\frac{r^{2} - 1}{2})$

$= 4 π k r^{2} - 4 π k$

$∴$ $q_{net} = q_{A} + q_{B} = 4 π k r^{2} - 3 π k = π k (4 r^{2} - 3)$

(1) $E_{net} = 0 \Rightarrow q_{net} = 0 \Rightarrow 4 r^{2} - 3 = 0$ $\Rightarrow r = \frac{\sqrt{3}}{2}$

So (1) is incorrect.

(2) $V = \frac{1}{4 π ε_{0}} \frac{q_{net}}{r} = \frac{1}{4 π ε_{0}} \frac{π k (4 r^{2} - 3)}{r}$

$\Rightarrow V = \frac{k}{4 ε_{0}} (4 r - \frac{3}{r})$ $\Rightarrow V = \frac{k}{4 ε_{0}} (4 \times \frac{3}{2} - \frac{3}{3} \times 2)$

$= \frac{k}{4 ε_{0}} (6 - 2) = \frac{k}{ε_{0}}$

So (2) is correct.

(3) If $r_{B} = 2,$ i.e. $r = 2$

then, $q_{net} = π k (16 - 3) = 13 π k .$

So (3) is incorrect.

(4) $E = \frac{k q_{net}}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{π k (4 r^{2} - 3)}{r^{2}}$ $= \frac{k}{4 ε_{0}} (4 - \frac{3}{r^{2}})$

$E = \frac{k}{4 ε_{0}} (4 - \frac{3}{25} \times 4)$ $= \frac{k}{4 ε_{0}} [\frac{88}{25}] = \frac{22 k}{25 ε_{0}}$

Hence (4) is correct.

Solution

1 If $r_{B} = \sqrt{\frac{3}{2}}$ , then the electric field is zero everywhere outside $B$ .

2 If $r_{B} = \frac{3}{2}$ , then the electric potential just outside $B$ is $\frac{k}{ε_{0}} .$

3 If $r_{B} = 2$ , then the total charge of the configuration is $15 π k .$

4 If $r_{B} = \frac{5}{2}$ , then the magnitude of the electric field just outside $B$ is $\frac{13 π k}{ε_{0}} .$

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Solution

1 If rB=32, then the electric field is zero everywhere outside B.

2 If rB=32, then the electric potential just outside B is kε0.

3 If rB=2, then the total charge of the configuration is 15πk.

4 If rB=52, then the magnitude of the electric field just outside B is 13πkε0.

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1 If $r_{B} = \sqrt{\frac{3}{2}}$ , then the electric field is zero everywhere outside $B$ .

2 If $r_{B} = \frac{3}{2}$ , then the electric potential just outside $B$ is $\frac{k}{ε_{0}} .$

3 If $r_{B} = 2$ , then the total charge of the configuration is $15 π k .$

4 If $r_{B} = \frac{5}{2}$ , then the magnitude of the electric field just outside $B$ is $\frac{13 π k}{ε_{0}} .$