Beam of particles in a cylindrical pipe

[CAF123]

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.

 

[tiny-tim]

Hi CAF123!

…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

 

[CAF123]

Hi CAF123! 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.

 

[tiny-tim]

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

 

[Nickie]

I would suggest double-checking your calculations and equations to ensure accuracy. It is possible that the solutions you checked have an error, but it is also possible that your calculations may have missed a step or made a mistake. It is important to verify your work and consider all factors before coming to a conclusion.

I would also suggest considering the physical principles behind the problem. In this case, the electric field is a vector quantity that points in the direction of the force on a positive charge. Therefore, the magnitude of the electric field at a certain point is dependent on the amount of charge enclosed by a Gaussian surface around that point. Make sure to take into account the charge distribution and the geometry of the problem when calculating the enclosed charge.

Additionally, it may be helpful to discuss your solution with a colleague or your instructor to get a second opinion and further clarify any discrepancies. Collaboration and discussion can often lead to a better understanding of the problem and its solution.