Calculating Electric Field Change due to UV Laser Emission

[02pedwsa]

Homework Statement

A very short pulse UV laser is used to liberate a number of electrons from the negative plate in the arrangement in (ii) above(describes parallel plate capacitor system, states plate thickness and separation in vacuum ignore edge effects, charge density and electric field calculated). The electrons drift towards the positive plate under the influence of the electric field over a time scale much longer than the laser pulse length. Suppose that after the centre of the cloud has traveled approximately 10 mm the spatial distribution of the charge is described by a Gaussian distribution with a standard deviation of 0.5 mm. Calculate the change in electric field across the electron cloud if 5×10^9 electrons m-2 were emitted during the laser pulse. Over what distance does the electric field effectively change?

Homework Equations

I believe i have to use gauss's law in its differential form V.E=P/e P=charge density e=8.85x10^12
with V.E being the divergence of the electric which i hope i can equate to the 10mm spatial distribution quoted but i really am stuck, I've been attempting it in all different ways and really haven't got a clue. If anyone can help please be as specific as possible as i could sit here all year otherwise

The Attempt at a Solution

ive made various attempts but all have thrown out wild looking answers

Cheers sam

 

[mark726]

Dear Sam,

Thank you for your post. I understand your confusion and I will try to explain the solution in a step-by-step manner. First, let's review the information given in the problem:

1. A UV laser is used to liberate electrons from the negative plate in a parallel plate capacitor system.
2. The electrons drift towards the positive plate under the influence of the electric field.
3. The time scale for the electron drift is much longer than the laser pulse length.
4. After traveling approximately 10 mm, the spatial distribution of the charge is described by a Gaussian distribution with a standard deviation of 0.5 mm.
5. 5×10^9 electrons m-2 were emitted during the laser pulse.

Now, let's start by calculating the charge density of the electron cloud. We can do this by using the number of electrons emitted and the area over which they are distributed. The area can be calculated by multiplying the plate thickness (given in the problem) by the distance traveled by the electron cloud (10 mm). This gives us an area of 0.05 m^2. Therefore, the charge density is equal to (5×10^9 electrons/m^2) x (1.6×10^-19 C/electron) = 8×10^-10 C/m^2.

Next, we can use Gauss's law in its differential form, V.E = P/ε, to calculate the electric field at any point in the cloud. Here, V is the electric potential, E is the electric field, P is the charge density, and ε is the permittivity of vacuum (8.85×10^-12 C^2/Nm^2).

We can simplify this equation by assuming that the electric field is constant throughout the cloud (which is not exactly true, but it will give us a good approximation). This assumption allows us to rewrite the equation as E = P/ε. Plugging in our values, we get E = (8×10^-10 C/m^2) / (8.85×10^-12 C^2/Nm^2) = 9.04×10^11 N/C.

Now, we can use the electric field to calculate the change in potential across the cloud. We can do this by using the equation V = Ed, where V is the potential difference, E is the electric field, and d is the distance traveled by the cloud (10 mm).