A long, solid, conducting cylinder has a radius of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{cm}}$. The electric field at the surface of the cylinder is$\mathbf{160}\mathbf{N}\mathbf{/}\mathbf{C}$, directed radially outward. Let A, B, and Cbe points that are$\mathbf{10}\mathit{c}\mathit{m}$,$\mathbf{ }\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{cm}}$, and$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{cm}}$, respectively, from the central axis of the cylinder. What are (a) the magnitude of the electric field at Cand the electric potential differences (b)${\mathbf{V}}_{\mathbf{B}}\mathbf{-}{\mathbf{V}}_{\mathbf{C}}$and (c)${\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}$?

The radius of the cylinder is,$R=2.0 \text{cm}=2.0\times {10}^{-2} \text{m}$.

The electric field at the surface of the cylinder is,$E_s=160N/C$

The separation between center of the cylinder and the point$A$is,$r_A=1.0 \text{cm}=1.0\times {10}^{-2} \text{m}$

The separation between center of the cylinder and the point$B$is,$r_B=2.0 \text{cm}=2.0\times {10}^{-2} \text{m}$

The separation between center of the cylinder and the point $C$is,$r_C=5.0 \text{cm}=5.0\times {10}^{-2} \text{m}$

After reading the question, Electric field at a distance $r$(outside) from the center due to charged cylinder of linear charge density$\lambda$is

$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$ …(1)

$V\left(r_2\right)-V\left(r_1\right)=-\underset{r_1}{\overset{r_2}{\int }}E\left(r\right)dr$ …(2)

Electric field at the surface of the cylinder$E_s$is expressed as,

$E_s=\frac{\lambda }{2\pi {\epsilon }_{0}R}$

From equation (1) electric field on the cylinder of radius$R$.

From equation (1) electric field outside the cylinder at a distance $r_C$from center is,

$E_o=\frac{\lambda }{2\pi {\epsilon }_0{r}_C}$

From the above two equations the electric field outside the cylinder is

$E_o=E_s\left(\frac{R}{r_c}\right)$

Substitute all the value in the above equation.

$$\begin{gathered} E_o=\left(160N/C\right)\left(\frac{2.0\times {10}^{-2}m}{5.0\times {10}^{-2}m}\right) \\ =64N/C \end{gathered}$$

Hence the magnitude of electric field at C is, $64N/C$.

Integrate the equation (2) limits from$r_B$to$r_C$to get potential difference between points$B$and$C$.

$$\begin{gathered} V_B-V_C={\int }_{r^n}^{r^c}{E}_s\left(\frac{R}{r}\right)dr \\ {V}_B-V_C=E_{s}R{\int }_{r^b}^{r^c}\frac{dr}{r} \end{gathered}$$

$V_B-V_C=E_{s}R\ln \left(\frac{r_C}{r_B}\right)$

Substitute all the value in the above equation.

$$\begin{gathered} V_B-V_C=\left(160N/C\right)\left(2.0\times {10}^{-2}m\right)ln\left(\frac{5.0\times {10}^{-2} \text{m}}{2.0\times {10}^{-2} \text{m}}\right) \\ =2.9 \text{V} \end{gathered}$$

Hence the potential difference between points B and C is, $29N$.

As the cylinder is conducting, the electric field inside the cylinder is zero. So the potential difference

$V_A-V_B=0$

Hence the potential difference between points A and B is, 0.