Question

The region between two concentric spheres of radii ‘a’ and ‘b’, respectively, has a volume charge density of $\rho ={\displaystyle \frac{A}{r}}$, where A is constant and r is the distance from the centre. At the centre of the sphere is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is:

A. ${\displaystyle \frac{Q}{2\pi a^2}}$
B. ${\displaystyle \frac{Q}{2\pi (b^2-a^2)}}$
C. ${\displaystyle \frac{Q}{2\pi (a^2-b^2)}}$
D. ${\displaystyle \frac{Q}{\pi a^2}}$

Answer 1

Hint: This question is an application of the concept of Flux that was given by Gauss. We can easily find the formula using Gauss's Theorem. Take a small element dr and then integrate. While integrating, set a and b as lower limit and upper limit respectively.

Complete answer:
Let us consider a sphere between two concentric spheres lying in the area of radius r and thickness dr. Charging within alone leads to electric field / flux according to Gauss' theorem.

So,
We have
$$\Rightarrow {\displaystyle \frac{kQ}{a^2}}={\displaystyle \frac{k[Q+\underset{a}{\overset{b}{\int }}4\pi r2drAr]}{b^2}}$$
Where, k is coulomb’s constant, and a, r, and b are the radii of the three concentric circles respectively.
Now,
On integrating, ‘rdr’
In accordance with the power rule of integration, i.e.,
$\int x^{a}dx={\displaystyle \frac{x^{a+1}}{a+1}}$
So,
We have
$$\Rightarrow Q{\displaystyle \frac{b^2}{a^2}}=Q+4\pi A[{\displaystyle \frac{r^2}{2}}{]}_a^b$$
$$=Q+4\pi A{\displaystyle \frac{(b^2-a^2)}{2}}$$
$$\Rightarrow {\displaystyle \frac{Q(b^2-a^2)}{a^2}}=2\pi A(b^2-a^2)$$
So, now
We have
$$A={\displaystyle \frac{Q}{2\pi a^2}}$$
So, the value of A such that the electric field in the region between the spheres will be constant, is $${\displaystyle \frac{Q}{2\pi a^2}}$$

So, the correct answer is “Option A”.

Note:
The law of Gauss, also known as the flux theorem of Gauss, is a law of physics relating to the propagation of electric charges to the resulting electric field. A closed one enclosing a volume such as a spherical surface might be the surface under consideration.