Electricity and Electric field

[Suyash Singh]

Homework Statement

Homework Equations

gauss law
q=charge on sphere
Q=total charge enclosed by gaussian surface
Q=alpha/r x (4/3 pi r^3-4/3 pi R^3) + q

The Attempt at a Solution

EA=Q/ε[/B]
E=Q/(Aε)
now
for E to be independent of r,

alpha/r x 4/3 pi r^3 + q = 1/(4)(pi)(r^2)
alpha x 4/3 pi r^2 +q= 1/(4)(pi)(r^2)
q=(4 alpha pi r^2)/3 - 1/(4)(pi)(r^2)

q=(16 alpha pi^2 r^4 - 3)/(4)(pi)(r^2)

but this is not even close to the answer

 

[kuruman]

You need to do an integral to find the total charge inside the Gaussian surface for $r > R$ because the charge density is not uniform in that region.

 

[Suyash Singh]

You need to do an integral to find the total charge inside the Gaussian surface for $r > R$ because the charge density is not uniform in that region.

but why we need to do that since we are only seeing the electric field intensity outside the sphere

 

[kuruman]

How would you find how much charge is in the region $R < r' <r$ where $r'$ is a radius between the surface of the sphere and the Gaussian surface of radius $r$? You cannot multiply the volume by the charge density, as you have done, because the amount of charge per unit volume decreases continuously as $r'$ increases.

 

[Suyash Singh]

Q=q+integral(R to r-R)[alpha/r 4/3 pi r^3]

but how do i get dr on side of integral?

 

[kuruman]

Q=q+integral(R to r-R)[alpha/r 4/3 pi r^3]

but how do i get dr on side of integral?

You started incorrectly. For volume charge density $\rho = \alpha/r$, $$Q=\int \frac{\alpha}{r}dV.$$
What is $dV$ in spherical coordinates? Note: This is a triple integral. If you don't remember, see "Integration and differentiation in spherical coordinates" here
https://en.wikipedia.org/wiki/Spherical_coordinate_system