JEE Main 2016 · 3 Apr · Shift 1 · Physics Q12

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PhysicsSingle correct+4 / -1Question 12 of 90

The region between two concentric spheres of radii ' $a$ ' and ' $b$ ', respectively (see figure), has volume charge density $\rho=\dfrac Ar$ , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :

Two concentric spheres of radii a and b with a point charge Q at the common centre.

\dfrac Q{2\pi a^2}

\dfrac Q{2\pi(b^2-a^2)}

\dfrac{2Q}{\pi(a^2-b^2)}

\dfrac{2Q}{\pi a^2}

Answer and worked solution

Official answer: A

Gauss's law at radius r:
$E\cdot4\pi r^2=\frac1{\epsilon_0}\left[Q+\int_a^r4\pi r'^2\frac A{r'}dr'\right]=\frac{Q+2\pi A(r^2-a^2)}{\epsilon_0}.$
E is independent of r when $Q=2\pi Aa^2$ , i.e. $A=\dfrac Q{2\pi a^2}$ .

Topics:Electric flux and Gauss's law

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 12