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
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 isBITSAT 2015
Solution
The problem involves the concept of electric field due to a uniformly charged solid sphere. The electric field E inside a uniformly charged solid sphere of total charge Q and radius R at a distance r from its center is given by E = kQr/R^3, where k is Coulomb's constant.
Given that the charge density ρ = kr^a, the total charge Q within a sphere of radius r is given by integrating the charge density over the volume of the sphere. This gives Q = ∫ρ dV = ∫kr^a dV = ∫kr^a 4πr^2 dr = 4πk ∫r^(a+2) dr = 4πk [r^(a+3)/(a+3)] from 0 to r = 4πk r^(a+3)/(a+3).
Substituting this into the expression for E gives E = k [4πk r^(a+3)/(a+3)] r/R^3 = 4πk^2 r^(a+4)/(a+3)R^3.
The problem states that the electric field at r = R/2 is 18 times that at r = R. Substituting these values into the expression for E gives 4πk^2 (R/2)^(a+4)/(a+3)R^3 = 18 * 4πk^2 R^(a+4)/(a+3)R^3.
Solving this equation for a gives a = 2. Therefore, the value of a is 2.
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