Question

Charge is distributed within a sphere of radius r with a volume charge density 2r / a 2 a (r) e r , where a and a are constants. If q is the total charge of this charge distribution, the radius r is

Answer 1

volume charge density of sphere is $\rho(r)=\frac{a}{r^2}e^{-\frac{2r}{a}}$

we know, volume charge density is the rate of charge distributed per unit volume.

so, $q=\int\limits^r_0{\rho(r)}\,dv$

Let r is the volume of sphere .

then, v = 4/3πr³

differentiating both sides,

dv = 4πr²dr, putting it in above expression.

so, $q=\int\limits^r_0{\frac{a}{r^2}e^{-\frac{2r}{a}}}4\pi r^2 dr$

= $4\pi a\int\limits^r_0{e^{-\frac{2r}{a}}}\,dr$

= $4\pi a^2/2\left[1-e^{-\frac{2r}{a}}\right]$

or, q/2πa² = 1 - e^{-2r/a}

or, 1 - q/2πa² = e^{-2r/a}

or, log(1 - q/2πa²) = -2r/a

or, 2r/a = log[1/(1 - q/2πa²)]

or, r = a/2 log[1/(1 - q/2πa²)]

hence, radius of sphere is a/2 log[1/(1 - q/2πa²)]

Answer 2

Answer:

In the file attached.

Explanation:

Thank you.