Point charges in a line; electric field

AI Thread Summary
The discussion focuses on determining the locations where the electric field is zero due to three point charges on the x-axis. One known location is between the charges at 0.2m and 0.4m, while the other is to the left of the first charge at 0m. Participants suggest setting up equations to sum the electric field vectors and emphasize the need to consider the direction and strength of the fields from each charge in different regions. A quartic equation can be derived to find the exact position of the zero electric field, with recommendations to plot the function for clarity. The conclusion is that the second zero point is indeed to the left of the origin, and finding its precise location requires careful calculation.
etagg
Messages
9
Reaction score
0

Homework Statement


Three point charges are on the x-axis at 0m, 0.2m, and 0.4m, with charges -19microC, +19microC, and +19microC respectively.

There are two places where the electric field is zero. One is between 0.2 and 0.4m. Where is the other place?


Homework Equations



i used the equation E=kq/r2


The Attempt at a Solution



I first recognized that the point must be to the left of the first charge.

Then, i tried to set up an equation summing the E field vectors:

(kq/r^2)-(kq/(r+0.2)^2)-(kq/(r+0.4)^2)=0
With this equation, i could not simplify to find a suitable x value.

I then tried to sum the E field due to the charges at 0.2m and 0.4m, and then use that to determine the distance r that the Electric field would be zero. I did this by recognizing that kq/(0.2)^2+kq/(0.4)^2=E', or the collective E field at 0m. I was then going to use E'=kq/x^2 because the fields are in opposite directions but have to be equal for the next field to be zero. This approach gave me the wrong answer.

Where am i going wrong?
 
Physics news on Phys.org
Hi etagg,

etagg said:

Homework Statement


Three point charges are on the x-axis at 0m, 0.2m, and 0.4m, with charges -19microC, +19microC, and +19microC respectively.

There are two places where the electric field is zero. One is between 0.2 and 0.4m. Where is the other place?


Homework Equations



i used the equation E=kq/r2


The Attempt at a Solution



I first recognized that the point must be to the left of the first charge.

Then, i tried to set up an equation summing the E field vectors:

(kq/r^2)-(kq/(r+0.2)^2)-(kq/(r+0.4)^2)=0
With this equation, i could not simplify to find a suitable x value.

I then tried to sum the E field due to the charges at 0.2m and 0.4m, and then use that to determine the distance r that the Electric field would be zero. I did this by recognizing that kq/(0.2)^2+kq/(0.4)^2=E', or the collective E field at 0m. I was then going to use E'=kq/x^2 because the fields are in opposite directions but have to be equal for the next field to be zero. This approach gave me the wrong answer.

Where am i going wrong?

I believe this is more of a conceptual problem. The idea is that there are four regions: to the right of the origin, between 0 and 0.2, between 0.2 and 0.4, and to the left of 0.4; and they have already said the region between 0.2 and 0.4 has a point where the E field is zero.

Now think about what needs to happen for the E field to be zero, and for each region, think about which way the fields are pointing from each charge, and how strong the field would be from each charge. You should find that only one region of the three that are left are capable of having a zero electric field from the three charges. Do you get the answer?
 
I think the electric field could only be zero to the left of the origin, as i stated in my solution attempt. I just don't know how to precisely find that place.
 
etagg said:
I think the electric field could only be zero to the left of the origin, as i stated in my solution attempt. I just don't know how to precisely find that place.

Oh, from the way the question was worded I though it was just asking for the region, not a particular x value.

To find the actual x value, I would use the first equation that is in your post. You can expand the parts out to get a quartic equation. At that point I would find the solution by plotting the function and seeing where it equals zero. (Or some calculators can solve it.)

(The way to solve a quartic by hand is a bit messy, so unless there is some simplification in this problem I'm not seeing, I think the other ways would be the best way to find the answer.)

If you do plot it, be sure to plot with r starting at zero and going to 1 (that is, don't let r be positive) because that is how you have written your function. Once you find the r, then you will let it be negative for the final answer.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top