Question

Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radius R = 7.5 cm. Assume that the powder and its charge were spread uniformly through the pipe with a volume charge density rho. (a) Find an expression for the electric potential as a function of the radial distance r from the center of the pipe. (The electric potential is zero on the grounded pipe wall.) (b) For the typical volume charge density rho = -3.7 × 10-3 C/m3, what is the difference in the electric potential between the pipe's center and its inside wall?

Answer 1

Answer:

A) $E =\frac{\rho r}{2\epsilon}$

B) $v = 58.7923\times 10^4 V$

Explanation:

a) using Guass law

$\oint E.dA = \frac{q_{enclosed}}{\epsilon_o}$

$EA = \frac{q_{enclosed}}{\epsilon_o}$

$E = \frac{q_{enclosed}}{A \epsilon_o}$

Here$A is = 2\pi rL$

$E = \frac{q_{enclosed}}{(2\pi rL) \epsilon_o}$

Volume charge density is given as

$\rho = \frac{q_{enclosed}}{volume}$

net charge is given as

$q_{enclosed} = \rho \times volume$

therefore $E = \frac{ \rho \times (L\pi r^2)}{(2\pi rL) \epsilon_o}$

$VOLUME = L\pi r^2$

After solving electric field equation we get

$E =\frac{\rho r}{2\epsilon}$

b) electric potential difference is given as

$v_{wall} - v = - \int_{r}^{R} Edr$

$0 - v = - \int_{r}^{R} E dr$

$v = \int_{r}^{R} Edr$

$= \int_{r}^{R} (\frac{\rho r}{2\epsilon}) dr$

$= \frac{\rho}{4\epsilon} (R^2 - r^2)$

at r = 0

$v = \frac{- 3.7 \times 10^{-3} \times 0.075^2}{4\times (8.85\times 10^{-12}}$

$v = 58.7923\times 10^4 V$