Electric Force acting on an electron

[format1998]

Homework Statement

A charge of -12.0nC is placed on the y-axis at y = -15cm and a charge of +12.0 nC is placed on the y-axis at y = 15cm. Determine (a) the electric field of each charge separately at x = 20.0cm, and (b) the net electric field due to these charges at x = 20.0 cm. (c) What force would act on an electron when it passes through this point on the x-axis moving at 1.73E6 m/s?

Homework Equations

E=k|Q|/r²
F=qE
c²=sqrt(a²+b²)

The Attempt at a Solution

(a) E=k|Q|/r² = [(8.99E9 Nm²/C²) |12E-9C|] / (0.25m)² = 1726 N/C

So...
E (due to +12 nC) = (1726 N/C , 323 degrees)
E (due to -12 nC) = (1726 N/C , 217 degrees)

(b) E_NET = 2*(1726 N/C)*cos(53 degrees) = 2077 N/C, 270 degrees

(c) my problem is this part
I am given v = 1.73E6 m/s but with F=qE_NET, velocity of the electron is irrelevant? or may be I'm wrong.

F=qE_NET=(1.60E-19 Nm²/C²)(2077 N/C , 270degrees)

According to the answer key, the direction of the Force is 90 degrees? Why? From my drawing attached, I'm seeing 270 degrees. What am I doing wrong?

 

[davo789]

You'll be pleased to know that your answers are all fairly good! As far as I can see, you are right in saying that the force on the electron is not at all related to its velocity, and you have calculated it correctly. I'll explain why the force is at 90 degrees in a bit of a long-winded way, but hopefully it will make sense!

You will find that calculations like this are much easier when you use vectors, rather than work in degrees all the time. I don't know what kind of level you're at, or whether you have even come across vectors before, but judging by the way you've drawn the diagram it looks like you have a reasonable knowledge of them!

The full formula for calculating the electric field is as follows:

$\underline{E}=\frac{q}{4\pi\epsilon_{0}|\underline{r}|^{2}} \hat{\underline{r}}$

Where $\hat{\underline{r}}$ is known as the 'unit vector'. This describes the direction in which the field points. Another thing to remember is the sign of q: positive charge = +q, negative = -q.

Calculating the unit vector is easy. We'll calculate it for your top charge first. $\hat{\underline{r}} = \frac{\underline{r}}{|\underline{r}|}$ In other words, you want to divide the vector pointing in the direction of the field by its magnitude. So, at x = 20cm, $\underline{r}=20x-15y$ (getting from your charge to that point), and $|\underline{r}|=\sqrt{20^{2}+15^{2}}$. After simplifying, $\hat{\underline{r}}=\frac{4x-3y}{5}$

I won't go through the calculation now, but the unit vector for the electric field caused by the lower charge is $\hat{\underline{r}}=\frac{4x+3y}{5}$ Can you see why?

The beauty with sorting things out this way is that adding the electric field becomes easy, since it is just a case of addition: $E_{NET}=\frac{q}{4\pi\epsilon_{0}|\underline{r}|^{2}} \frac{4x-3y}{5} + \frac{-q}{4\pi\epsilon_{0}|\underline{r}|^{2}} \frac{4x+3y}{5} = \frac{q}{4\pi\epsilon_{0}|\underline{r}|^{2}} (\frac{-6y}{5}) = -2071 y N/C$ In other words, we've just shown that the E field has a magnitude of 2071N/C (close enough to yours!) and is pointing straight down the y axis, which is what you also showed, except you described its direction with '270 degrees'.

To answer your initial confusion as to why the electron experiences a force 'at 90 degrees' as you put it, $F = qE = -q * -2071 y = 2071q y$. The minus sign flips the direction to the +ve y direction. This is really easy to do when you use the notation that I have, but a bit more tricky if you're working in degrees.

I really hope this has helped a little! Let me know if you want me to word things again or stop writing long-winded answers.

 

[format1998]

Hello davo789!

Thank you for the explanation. All of it does make sense. It's just that my professor uses the degree "method", so I try to use what he does. And another one of my problems is that I've been out of school for the last 4 years and this is my semester back. I do remember simple vector calculations but the more complicated ones I have forgotten. I am in the process of relearning them.

Again, thank you very much for your help. I do appreciate any and all explanations, long winded or not. I appreciate the time you took to fully explain everything. Maybe you can take a look at my other post titled "Electric Field" https://www.physicsforums.com/showthread.php?t=527093. As I am still having so much trouble with that one. I know my problem is vector calculations but I just don't know what I'm doing wrong... Thank you

~ format1998

 

[davo789]

No problem! As you say, it's probably smart to stick with what your lecturer is doing, for now at least!

 

[rwisz]

I would like to provide a clarification on the direction of the force acting on the electron in this scenario. The direction of the force can be determined using the right hand rule, where the thumb points in the direction of the velocity of the electron, the index finger points in the direction of the magnetic field (which is zero in this case), and the middle finger points in the direction of the force. In this case, the direction of the force would be perpendicular to the velocity of the electron, which is why the correct answer is 90 degrees.