I think there are mistakes in the concept that I'm applying -- Charged sphere's potential

[Neolight]

Homework Statement

A sphere of radius R has a volume charge density proportional to distance from center. Total charge contained in the sphere is Q. If electric potential at infinity is taken to be zero , potential at it's center is

Homework Equations

Here given are
ρ= βr where β is a constant

The Attempt at a Solution

Since in this question the charger density is not uniform so I attempted the question by first finding out the electric field on the inside and outside region by using Gauss law than.

Please give me some hints on what I have done wrong because the answer is
V= Q/3Πξ.R
And I also think that somehow we have to find the value of β

And one final personal question
Is this question a difficult level question because I have spent like 2 hours in a day for 4 days and haven't been able to answer due to mistakes here and there and lately because of this I have started losing confidence in myself because I can't even solve this question.

 

[Charles Link]

I don't want to read your sideways picture, but $Q=\int\limits_{0}^{R} \beta r \, 4 \pi r^2 dr$. That will allow you to find $\beta$. You then need to find $E(r)$, (by Gauss' law), as a function of $r$ both inside and outside the sphere. That will allow you to compute $V(r)=-\int\limits_{+\infty}^{r} E(r') \, dr'$. $\\$ Maybe I'm a little fussy, but I think it is asking a lot to have someone need to turn their computer sideways to try and read a sideways picture.

 

[Neolight]

I'm really sorry about that tilted photos.
Here are my calculations

 

[Charles Link]

It looks quite accurate. The equation $\beta \pi R^4=Q$ applies, so that $\beta=Q/( \pi R^4)$. You basically computed this, but didn't include it. You can substitute this for $\beta$ in your final answer. From what I can tell it looks right=I didn't see any mistakes.

 

[Ray Vickson]

View attachment 214150 View attachment 214151

I'm really sorry about that tilted photos.
Here are my calculations

You say your answer of V= Q/3Πξ.R is wrong. I cannot figure out if you mean $\frac{Q}{2 \pi \xi R}$ or $\frac{Q}{3 \pi \xi} R$ or something else. Parentheses would help.

Anyway, why do you think you answer is wrong? What is the apparent correct answer?

(I, for one, will not look at posted images; I look at typed work only, except for diagrams.)

 

[Charles Link]

You say your answer of V= Q/3Πξ.R is wrong. I cannot figure out if you mean $\frac{Q}{2 \pi \xi R}$ or $\frac{Q}{3 \pi \xi} R$ or something else. Parentheses would help.

Anyway, why do you think you answer is wrong? What is the apparent correct answer?

(I, for one, will not look at posted images; I look at typed work only, except for diagrams.)

Looks like the OP got the right answer, which is $V=\frac{Q}{3 \pi \epsilon_o R}$. And the OP should use an $\epsilon_o$ (epsilon), rather than a $\xi_o$ (xi).