Answer:
The electric field at a distance r = 0.02 m is 14062.5 N/C.
Solution:
Refer to fig 1.
As per the question:
Radius of sphere, R = 0.04 m
Charge, Q = [tex]5.0\times 10^{- 9} C[/tex]
Distance from the center at which electric field is to be calculated, r = 0.02 m
Now,
According to Gauss' law:
[tex]E.dx = \frac{Q_{enclosed}}{\epsilon_{o}}[/tex]
Now, the charge enclosed at a distance r is given by volume charge density:
[tex]\rho = \frac{Q_{enclosed}}{area}[/tex]
[tex]\rho = \frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}}[/tex]
Also, the charge enclosed Q' at a distance r is given by volume charge density:
[tex]\rho = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]
Since, the sphere is no-conducting, Volume charge density will be constant:
Thus
[tex]\frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}} = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]
Thus charge enclosed at r:
[tex]Q'_{enclosed} = \frac{Q_{enclosed}}{\frac{r^{3}}{R^{3}}[/tex]
Now, By using Gauss' Law, Electric field at r is given by:
[tex]4\pi r^{2}E = \frac{Q_{enclosed}r^{3}}{\epsilon_{o}R^{3}}[/tex]
Thus
[tex]E = \frac{Q_{enclosed}r}{4\pi\epsilon_{o}R^{3}}[/tex]
[tex]E = \frac{(9\times 10^{9})\times 5.0\times 10^{- 9}\times 0.02}{0.04^{3}}[/tex]
E = 14062.5 N/C
