JEE Advanced 2022 · Paper 2 · Physics Q9

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PhysicsMultiple correct+4 / -2Question 27 of 54

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 $\rho_A = kr$ , where $k$ is positive. In the spherical shell $B$ of outer radius $r_B$ , the electrostatic charge density varies as $\rho_B = \dfrac{2k}{r}$ . Assume that dimensions are taken care of. All physical quantities are in their SI units.

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

Two concentric circles: an inner shaded sphere A of radius rA, surrounded by a spherical shell B extending from rA out to radius rB.

r_B = \sqrt{\dfrac32}

, then the electric field is zero everywhere outside

B

r_B = \dfrac32

, then the electric potential just outside

B

\dfrac{k}{\epsilon_0}

r_B = 2

, then the total charge of the configuration is

15\pi k

r_B = \dfrac52

, then the magnitude of the electric field just outside

B

\dfrac{13\pi k}{\epsilon_0}

Answer and worked solution

Official answer: B

Charges.

Sphere A: $Q_A=\int_0^14\pi r^2(kr)\,dr=\pi k$ .
Shell B: $Q_B=\int_1^{r_B}4\pi r^2\frac{2k}r\,dr=4\pi k(r_B^2-1)$ .

Total: $Q=\pi k(4r_B^2-3)$ .

Check each option.

(A) $r_B^2=\frac32$ : $Q=3\pi k\ne0$ , so the field outside is not zero. False.
(B) $r_B=\frac32$ : $Q=6\pi k$ , and $V=\dfrac{Q}{4\pi\epsilon_0r_B}=\dfrac{6\pi k}{6\pi\epsilon_0}=\dfrac k{\epsilon_0}$ . True.
(C) $r_B=2$ : $Q=13\pi k$ , not $15\pi k$ . False.
(D) $r_B=\frac52$ : $Q=22\pi k$ , so $E=\dfrac{22\pi k}{4\pi\epsilon_0(25/4)}=\dfrac{22k}{25\epsilon_0}$ , not $\dfrac{13\pi k}{\epsilon_0}$ . False.

(B).

Topics:Electric flux and Gauss's law Electric potential

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JEE Advanced 2022 · Paper 2, Physics question 9