Question

The region between two concentric spheres of radii ‘a’ and ‘b’, respectively, has volume charge density ρ=A/r, 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:

(1) Q/[2π(b2-a2)]

(2) 2Q/[π(b2-a2)]

(3) 2Q/πa2

(4) Q/2πa2

...... Answer this with reason..........​

Answer 1

Dear Friend

Please find below the solution to the asked query:

According to Gauss's Law, if we consider the gaussian surface with radius a, then the electric field at that point is,

E.(4πa2)=Qε0⇒E=Q4πε0a2

Now, if we consider the gaussian surface with the radius b, then the electric field at that point is,

E.(4πb2)=Q+∫baAr4πr2drε0⇒E=Q+4πA∫bardr4πε0b2⇒E=Q+2πA(b2−a2)4πε0b2

Given that the electric field should be constant. Therefore,

⇒Q+2πA(b2−a2)4πε0b2=Q4πε0a2⇒Qb2+2πA(b2−a2)b2=Qa2⇒2πA(b2−a2)b2=Qa2−Qb2⇒2πA(b2−a2)b2=Q(b2−a2)a2b2⇒2πA=Qa2⇒A=Q2π a2

Hope this information will clear your doubts about the topic.

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Answer 2

Let us find total charge enclosed in a sphere of radius r,

Q ′ =Q+∫ ar rA 4πr 2 dr=Q+2πAr 2 −2πAa 2

By Gauss law,

E×4πr 2 =Q−2πAa 2 +2πAr 2

Given, E is independent of r

Hence, Q−2πAa 2 =0

This gives A= Q/2πa²