Electrostatic Potential inside sphere of uniform charge.

[jspectral]

1. Given a sphere of uniform charge Q, radius R: Find an expression for the electrostatic potential V as a function of r for r ≤ R.
Prior proof: E= Q(r)/(4πϵ₀)r^2 ,for r≤R where Q(r) is the excess charge in the spherical volume of radius r

2. Relevant equation: V = Q/(4πϵ₀)r

3. I tried integrating for spheres of radius r_i over 0 - R where ∆Q_i is each small bit of charge and ∆Q_i = (Q/V)∆V_i giving (3Qr²/R³)∆r.
My eventual answer resulted in (Q/4πϵ₀)(R³ - r³)

I know there's some form of integral involved, but I'm not sure what/where.
The answer (shown on the answers) is shown to be V (r) = [Q/(8πϵ₀R³)](3R² - r²) but no working is provided, and I can't for the life of me figure out what they did. Thanks.

 

[lanedance]

have a look at this one
https://www.physicsforums.com/showthread.php?t=352765

 

[jspectral]

have a look at this one
https://www.physicsforums.com/showthread.php?t=352765

I tried this method:
$$V(r) - V(R) = \int_{R}^r E.dr$$ where $$E = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2}$$

But I only end up with factors of $$\frac{1}{R}$$ and $$\frac{1}{r}$$

Am I approaching the problem wrong? Should I be integrating over a different domain?

Considering the answer: $$V(r) = \frac{Q}{8\pi\epsilon_{0}R^3}(3R^2 - r^2)$$
I don't get which integral could give the difference in scalar in front of r and R, nor where the bottom factor of $$R^3$$ comes from?

I also tried:
$$\eta = \frac{Q}{V} = \frac{3Q}{4\pi R^3} \Rightarrow \Delta Q_{i} = \frac{3Q}{4\pi R^3}\Delta V_{i} = \frac{3Q}{4\pi R^3}4\pi r^2\Delta r$$

$$\Rightarrow V(r) = \frac{3Q}{4\pi\epsilon_{0}R^3}\int_{R}^r r^2 dr$$

Which is closer, but still evades the factor of 3 infront of R, and the factor of 8 on the bottom, and the powers of r and R are incorrect. =S