Electric Field Due to a Point Charge

AI Thread Summary
The discussion revolves around calculating the coordinate on the x-axis where the net electric field from two fixed charges is zero. The charges are q1 = 2.1 * 10^-8 C and q2 = -4.00q1, positioned at x = 70 cm. The user attempts to set up equations based on the electric field equations but struggles to isolate the variables. A suggestion is made to relate the distances r1 and r2 using the total distance of 70 cm, which helps clarify the problem. The conversation emphasizes the importance of correctly applying the equations and checking for errors in calculations.
Wables
Messages
5
Reaction score
0

Homework Statement


Two particles are fixed to an x axis: particle 1 of charge q1= 2.1 * 10^-8 and particle 2 of charge q2= -4.00q1 at x=70 cm. At what coordinate on the axis is the net electric field produced by the particles equal to zero?

Homework Equations



Enet = E1 + E2
E=F/q
F=k q1q/r2
Let q = point charge in the above formulas

The Attempt at a Solution



Enet = E1 + E2 = 0
E1 = -E2

k q1/r12 = -k q2/r22
q1/r12 = q2/r22

At this point I thought I should try to isolate for one of the r values and plug it into a second equation to solve for one of the r values. But there is no other equation I can use to set up a system of equations... or is there? I am stuck here.

All help appreciated,
Thanks!
 
Last edited:
Physics news on Phys.org
Welcome to PF!

Hi Wables! Welcome to PF! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
Wables said:
At this point I thought I should try to isolate for one of the r values and plug it into a second equation to solve for one of the r values. But there is no other equation I can use to set up a system of equations... or is there?

you need an equation relating r1 and r2 :wink:
 
use r1 + r2 = x = 70 cm
 
Thanks for the welcome.

r1+x=20
r2+x=70

I can use these! Ooooh!

Okay, so:


q1/r12=q2/r22
=> (q1/q2)(1/2)r2=r1

Let (q1/q2)(1/2) = z

r2+x=70
=> r2=70-x

r1+x=20
=> zr2 + x = 20
=> z(70-x) + x = 20
=> x-zx = 20 - 70z

But z is a root of a negative number...?
 
Hi Wables! :smile:
Wables said:
q1/r12=q2/r22

No.

Did you draw a diagram?

You need a minus in there, don't you? :wink:
 
OH! Haha thanks for catching that. Okay. Thanks for your help, this question was really bugging me! :)
 
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 '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