Respuesta :
Answer:
Electric field, [tex]E = \frac{a}{2epsilon}[/tex]
Given:
charge density, [tex]\rho = \frac{a}{r}[/tex]
a = constant
Solution:
Area of sphere = [tex]4\pi r_{2} dr[/tex]
Small charge dq on a sphere of thickness r is:
dq = [tex]\rho \times A\times dr[/tex]
dq = [tex]4\rho \pi r_{2} dr[/tex] (1)
Integrating both the sides, we get:
[tex]dq = 4\rho \pi r_{2} dr[/tex]
[tex]\int dq = \int \frac{a}{r}\times 4\pi r^{2} dr[/tex]
[tex]q(r)= 2\pi r^{2}a - 0 = 2\pi r^{2}a[/tex]
Therefore, the total charge enclosed by a sphere of radius r is [tex]q(r)= 2\pi r^{2}a[/tex]
Now, by Gauss Law, flux emerging out of the surface is equal to [tex]\frac{1}{\epsilon_{o}}[/tex] times the total charge enclosed:
[tex]A\times E = \frac{q}{\epsilon_{o}}[/tex]
[tex]4\pi r^{2}\times E = \frac{}{\epsilon_{o}}[/tex]
[tex]E = \frac{2\pi r^{2}a}{\epsilon_{o}}[/tex]
[tex]E = \frac{a}{2epsilon}[/tex]
where
E = electric field
