Assuming that the electric field at a distance $r$ from the center of a non-conducting sphere with radius $R$ and uniformly distributed charge $Q$ is $E=\frac{1}{4\pi\epsilon_0}\frac{Q}{R^3}r$, we are asked to find the electric potential at a distance $r$ away from the center.
My work for this problem is as follows:
$\Delta V=V_r-V_0=\frac{-W_r}{q}-\frac{W_0}{q}=\frac{-Wr}{q}=\frac{-Fd}{q}=\frac{-qEd}{q}=-Ed=\frac{1}{4\pi\epsilon_0}\frac{Qrd}{R^3}$. ($W_0=0$ since the electric field at radius $0$ is $0$).
However, the answer is (suposedly) $V=\frac{Q}{4\pi\epsilon_0}\left(\frac{3R^2-r^2}{2R^3}\right)$. Why is my work/approach incorrect?
Best Answer
By definition, the potential difference between two separate points A and B is $$ V_{BA} := - \int_A^B \vec{E}\cdot d\vec{r}. $$ (Note that you can only use the result $V_{BA} = |\vec{E}|d_{BA} =|\vec{F}|d_{BA}/q$ when you have an electric field that is constant between the two points. In this case it is not so you have to use the integral definition.)
While it is unambiguous to describe a potential difference between two points, the potential at a point can be described with respect to some reference point. This reference point is arbitrary but it is often taken at infinity where many potentials are defined to be zero. It seems that this is the case here.
The potential is $$ V = - \int_{\infty}^r \vec{E}\cdot d\vec{r'}. $$ Because the electric fields are spherically symmetric, the integral can be reduced to the 1D version $$ V = - \int_{\infty}^r E(r')\,dr'. $$
However for this problem, the form of the electric field is different for $r'>R$ and $r'<R$. Split the integral for these different conditions and you'll find the answer.