IIT JEE Main 2016 Physics Question No 10

IIT JEE Main 2016 Physics Question No 10

The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure ), have 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:

IIT JEE Main 2016 Physics Question No 10
(1) 2Q/(Π(a2-b2))
(2) 2Q/(Π(α2)
(3) Q/(2Π(α2)
(4) Q/(2Π(b2-a2))

Solution

Using Gauss Theorem for a Gaussian surface for radius r, we have

E 4Πr2 = (1/ε0)[Q + ∫ar(A/r)4Πr2dr]
E = (Q/(4Πε0r2)) + [(1/(4Πε0r2))2ΠA{r2 – ar2}]
E = (Q/(4Πε0r2)) + (A/2ε0) – (Aa2/2r2ε0)

for E to be independent of r

(Q/(4Πε0r2)) – (Aa2/2r2ε0) = 0

=> A = Q/2Πa2

Answer

Correct answer to given problem No 10 of IIT JEE Main Physics paper of 2016 is A = Q/2Πa2 i.e. option number 3.

Save

Save

Save