Question

Charge is uniformly distributed with charge density \rho inside Calc a very long cylinder of radius R. Find the potential difference between the surface and the axis of the cylinder.

Short answer

The potential difference between the surface and axis of the cylinder

$=-\frac{\rho m^3}{4e_y}$

Step-by-step solution

given information

Given Info:

Charge is uniformly distributed with charge density \rho inside a very long cylinder of radius R.

Formula Used:

If $\overrightarrow{E}$represents electric field and V represents potential associated with it then they are related by $vecE=-\nabla V$

Gauss's law, E. $A=\frac{Q_m}{{ϵ}_0}.$

Where E$\to$

electric field through the surface of area A

$Q_{mu}\to$ charge enclosed by the suiface

${ϵ}_0$$\to$ Permittivity of vacuum

calculation

Consider a Gaussian surface inside the cylinder with radius r

Applying Gauss's law to the Gaussian surface,

$-E\times 2\pi rl=\frac{\pi r^{2}lp}{{ϵ}_9}$

gives$E=\frac{m}{2e_i}$

So potential difference between the surface and axis of the cylinder is given by,

$\Delta V=-{\int }_0^{R}Edr=-{\int }_0^R\frac{\rho r}{2{ϵ}_{\Vert }}dr$

Therefore,$\Delta V=-\frac{p}{2{ϵ}_u}{\left[\frac{r^2}{2}\right]}_{11}^R=-\frac{a{N}^2}{4{ϵ}_p}.$