## 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:

(1) 2Q/(Π(a^{2}-b^{2}))

(2) 2Q/(Π(α^{2})

(3) Q/(2Π(α^{2})

(4) Q/(2Π(b^{2-}a^{2}))

#### Solution

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

E 4Πr^{2} = (1/ε_{0})[Q + ∫_{a}^{r}(A/r)4Πr^{2}dr]

E = (Q/(4Πε_{0}r^{2})) + [(1/(4Πε_{0}r^{2}))2ΠA{r^{2} – ar^{2}}]

E = (Q/(4Πε_{0}r^{2})) + (A/2ε_{0}) – (Aa^{2}/2r^{2}ε_{0})

for E to be independent of r

(Q/(4Πε_{0}r^{2})) – (Aa^{2}/2r^{2}ε_{0}) = 0

=> A = Q/2Πa^{2}

#### Answer

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