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

A)
`E =(\rho r)/(2\epsilon)`

B)
`v = 58.7923* 10^4 V`

Step-by-step explanation:

a) using Guass law

`\oint E.dA = (q_(enclosed))/(\epsilon_o)`

`EA = (q_(enclosed))/(\epsilon_o)`

`E =  (q_(enclosed))/(A \epsilon_o)`

Here
`A is = 2\pi rL`

`E =  (q_(enclosed))/((2\pi rL) \epsilon_o)`

Volume charge density is given as

`\rho = (q_(enclosed))/(volume)`

net charge is given as

`q_(enclosed) = \rho * volume`

therefore
`E =  ( \rho * (L\pi r^2))/((2\pi rL) \epsilon_o)`

`VOLUME  =  L\pi r^2`

After solving electric field equation we get

`E =(\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) ((\rho r)/(2\epsilon)) dr`

`= (\rho)/(4\epsilon) (R^2 - r^2)`

at r = 0

`v = (- 3.7 * 10^(-3) * 0.075^2)/(4* (8.85* 10^(-12))`

`v = 58.7923* 10^4 V`