Question

Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by r(r) = kr, where r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then :
(A) a = 3R/(2)¹/⁴
(B) a = (2)⁻¹/⁴.R
(C) a = (8)⁻¹/⁴.R
(D) a = R /√3

Answer 1

Given: Total charge 2Q is distributed in a sphere of radius R, density given by r(r) = kr.

To find: The value of a?

Solution:

• Now we have given the density function as:

              r(r) = kr

• We have also given that two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance, a, from the centre.

• So, we have:

              E x 4πa² = ∫ kr4πr² dr / ∈o      (here integration limit is from 0 to 1 )

              E = k x 4πa^4 / 4 x 4π∈o  

              2Q = ∫ kr x 4πr² dr                   (here integration limit is from 0 to R )

              k = 2Q / πR^4  ..........(i)

              Q = kπR^4 / 2

• Now QE = kQ²/(2a)²   .........(ii)

• Now adding (i) and (ii), we get:

              R = a x 8^1/4

             a = R/8^1/4 = R x 8^-1/4

Answer:

          So the value of a is R x 8^-1/4.