A solid sphere of radius R𝑅 has a charge Q𝑄 distributed in its volume with a charge density ρ𝜌 =kra=𝑘𝑟𝑎, where k𝑘 and a𝑎 are constants and r𝑟 is the distance from its centre. If the electric field at r=𝑟= R2𝑅2 is 1818 times that at r=R,𝑟=𝑅, the value of a𝑎 is
Question
A solid sphere of radius R𝑅 has a charge Q𝑄 distributed in its volume with a charge density ρ𝜌 =kra=𝑘𝑟𝑎, where k𝑘 and a𝑎 are constants and r𝑟 is the distance from its centre. If the electric field at r=𝑟= R2𝑅2 is 1818 times that at r=R,𝑟=𝑅, the value of a𝑎 is
Solution
The electric field E inside a uniformly charged solid sphere of total charge Q and radius R at a distance r from the center is given by E = kQr/R^3, where k is Coulomb's constant.
However, in this case, the charge density ρ is not uniform but varies with r as ρ = kr^a. The total charge enclosed within a sphere of radius r is then given by the integral of ρ over the volume of the sphere, which is Q = ∫ρ dV = ∫kr^a dV from 0 to r.
The differential volume element in spherical coordinates is dV = 4πr^2 dr, so Q = ∫4πk r^(a+2) dr from 0 to r = 4πk/(a+3) r^(a+3) for a ≠ -3.
The electric field at a point inside the sphere is then given by Gauss's law as E = kQ/r^2 = k * 4πk/(a+3) r^(a+3) / r^2 = 4πk^2/(a+3) r^(a+1) for a ≠ -3.
Given that the electric field at r = R/2 is 18 times that at r = R, we have 4πk^2/(a+3) (R/2)^(a+1) = 18 * 4πk^2/(a+3) R^(a+1).
Solving this equation for a gives a = -1. Therefore, the value of a is -1.
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A solid sphere of radius R𝑅 has a charge Q𝑄 distributed in its volume with a charge density ρ𝜌 =kra=𝑘𝑟𝑎, where k𝑘 and a𝑎 are constants and r𝑟 is the distance from its centre. If the electric field at r=𝑟= R2𝑅2 is 1818 times that at r=R,𝑟=𝑅, the value of a𝑎 isBITSAT 2015
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