Integrating dq to find that q(r) = Q(1-e^(-r/R))

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To find the charge q(r) enclosed in a sphere of radius r, the integral dq = rho(r) * 4πr^2dr must be evaluated, where rho(r) is given as Q * e^(-r/R) / (4πRr^2). The integration should be performed in spherical coordinates, using the volume element d^3r. The integration will lead to the expression q(r) = Q(1 - e^(-r/R)). It's emphasized that understanding the integration process is crucial for arriving at the correct result. Properly setting up the integral is key to solving the problem effectively.
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Homework Statement


provided with data that
dq = rho(r) *4phi*r^2*dr
rho(r) = [Q*e^(-r/R) / 4phi R *r^2)

I have to show that the charge q(r) enclosed in a sphere of radius r is q(r) = Q(1-e^(-r/R)) by using appropriate integral. how the integral should be?

Homework Equations





The Attempt at a Solution


I've tried to integrate dq = ... but I can't find the final answer that q(r) = Q(1-e^(-r/R))
 
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jk0921 said:
I've tried to integrate dq = ... but I can't find the final answer that q(r) = Q(1-e^(-r/R))

It is a pretty straightforward integration. Why not post what you've tried so we can see where you are going wrong?
 
never mind
 
The easiest way is to integrate the charge density in a fitted coordinate system!

Cause you need
Q(r) = \int \limits_{\mathcal{V}} \, d^3r \, \rho(r)​
of a sphere, the most suitable one is the spherical coordinate system. So you need the volume element
d^3r = \rm{?}​
and perform the integration!



PS:
In the statement
dq = 4\pi \, r^2 \,\rho(r) \cdot dr​
two integrations are already perfomed, so the best way to undestand it completley is to do it like I've said above!
 
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