91Ó°ÊÓ

We have given that the potential on the surface of the cylinder is $V0$, and the electric field outside the cylinder is $Er=1/2pP0r$.

We need to find the the potential relative to the surface at a point that is distance $r$from the axis, assuming $r7R$

The potential difference is given as in relation to the electric field strength and the displacement as

$∆V=-Edl$

where, $dl$is the path. Only we have one dimension, the length from the cylinder, which means that we only need to consider the parameter $r$.That is,

$∆V\left(r\right)=-{∫}_R^{r}E\left(r\right)dr$

The simple formula for potential difference is

$∆V\left(r\right)=V\left(r\right)-V\left(R\right)$

Therefore, the potential we have to find is

$V\left(r\right)=∆V\left(r\right)+V\left(R\right)$,

which, after substituting the expressions we know, becomes

$V\left(r\right)=-{∫}_R^{r}E\left(r\right)dr+V_0$.

Solving the integration part

${∫}_R^{r}E\left(r\right)dr={∫}_R^r\frac{\mathrm{λ}}{2\mathrm{Ï€}{∈}_0}dr=\frac{\mathrm{λ}}{2\mathrm{Ï€}{∈}_0}{∫}_R^r\frac{dr}{r}=\frac{\mathrm{λ}}{2\mathrm{Ï€}{∈}_0}{\left[In\right|r\left|\right]}_R^r==\frac{\mathrm{λ}}{2\mathrm{Ï€}∈}$₀(Inr-InR)=λ2π∈0InrR

Hence, the potential is

$V\left(r\right)=V_0-\frac{\mathrm{λ}}{2\mathrm{π}{∈}_0}In\frac{r}{R}$