IIT-JEE 2009 · Paper 2 · Physics Q56

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PhysicsInteger+4 / -1Question 56 of 57

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=\kappa r^a$ , where $\kappa$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r=\dfrac R2$ is $\dfrac18$ times that at $r=R$ , find the value of $a$ .

Your answer

Answer and worked solution

Official answer: 2

Enclosed charge. $q(r)=\displaystyle\int_0^r\kappa r'^a4\pi r'^2dr'\propto r^{a+3}$ , so inside the sphere $E\propto\dfrac{q(r)}{r^2}\propto r^{a+1}$ .

Condition. $\left(\frac12\right)^{a+1}=\frac18$ , so $a+1=3$ and $a=2$ .

Answer: 2

Topics:Electric flux and Gauss's law

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IIT-JEE 2009 · Paper 2, Physics question 56