Beam of particles in a cylindrical pipe

AI Thread Summary
The discussion revolves around calculating the electric field of charged particles in a cylindrical beam using Gauss's Law. The participants analyze the charge density and the enclosed charge within a Gaussian surface, leading to different interpretations of the charge calculations. One participant believes the provided solution for the enclosed charge is dimensionally incorrect, while others clarify that the charge density is indeed nq per unit volume. The conversation highlights the importance of accurately determining the charge within the Gaussian cylinder to derive the correct electric field expression. Ultimately, the accuracy of the calculations is crucial for understanding the behavior of the electric field in this context.
CAF123
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Homework Statement


Charged particles, each holding charge q are moving in a cylinderical beam centred on the x-axis with n particles per unit volume. All the particles have the same horizontal velocity v.

A) By considering a suitable Gaussian surface, calculate the E-field as a function of r, the radial distance from the x-axis, and hence the force on the charges particle due to the electric field.

2. Homework Equations

Gauss Law,

The Attempt at a Solution



Let a be the radius of the pipe. Choose a Gaussian cylinder to be of radius r < a. Then the E field (from the enclosed charge) and the dA elements are parallel, so by Gauss,## E∫dA = Q_{enc}/ε = E(2 \pi r h),## h the height of the pipe and Gaussian cylinder.

I then said that the volume charge density is ##Q/\pi a^2 h##. So in the Gaussian cylinder, the charge enclosed is ##(\pi r^2 h) \cdot Q/\pi a^2 h = \left(\frac{r}{a}\right)^2 nq## which then gives me the E field and hence the force. My problem is, when I checked the solutions, they say the charge enclosed is ##Q = nq \pi r^2 h## and then they get an E field of ##nrq/2\epsilon##. To be honest, I think this is wrong. This expression for Q yields incorrect dimensions and then when they calculate the E field, they have ##Nm^3/C## which again is wrong. Both my expressions give the correct dimensions. Am I correct?

Many thanks.
 
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Hi CAF123! :smile:
CAF123 said:
…with n particles per unit volume.

I then said that the volume charge density is ##Q/\pi a^2 h##. So in the Gaussian cylinder, the charge enclosed is ##(\pi r^2 h) \cdot Q/\pi a^2 h = \left(\frac{r}{a}\right)^2 nq##…

I don't understand …

q is charge, n is 1/volume
 
tiny-tim said:
Hi CAF123! :smile:I don't understand …

q is charge, n is 1/volume

Yes, there are n particles per unit volume so charge of nq per unit volume. So (volume) charge density is ##nq/(\pi a^2 h)##. Then I multipled this by the volume of the Gaussian cylinder to get the charge within the Gaussian cylinder.
 
CAF123 said:
Yes, there are n particles per unit volume so charge of nq per unit volume. So (volume) charge density is ##nq/(\pi a^2 h)##.

No, there's a charge of nq per m3.

Volume of cylinder = πa2h m3, so total charge in cylinder = πa2hnq,
and charge inside radius r = πr2hnq
 
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