Point Charges, Maginitudes and Electric Fields.

  • #1
Northbysouth
249
2

Homework Statement



Point charge 2.5 microC is located at x = 0, y = 0.30 m, point charge -2.5 microC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.5 microC at x = 0.40 m, y = 0?

Homework Equations


pythagorean theorem

E = kq/r^2
F = qE =


The Attempt at a Solution



I drew the charges on an x-y graph. Then I tried to find the electric field of a single charge:

E1 = (9x10^9 Nm^2/C)(2.5x10^-6 C)(4.5x10^-6 C)/(0.5^2 m)(0.4/0.5))

E1 = 0.50625 m

I then doubled this value to get 1.0125 m. Which is says is wrong. Any suggestions?
 

Answers and Replies

  • #2
voko
6,054
391
Why do you think you can just double the value?
 
  • #3
Northbysouth
249
2
Because there's two point charges.
 
  • #4
Muphrid
834
2
Vectors, man. Doing what you did is like saying you can add 1 mile north and 1 mile east and get 2. It's nonsense.
 
  • #5
Northbysouth
249
2
I've managed to get the right answer but I don't quite understand all of it. Here's what I did:

I labelled the point at (0, 0.30) as charge 1. I then found the force of this point using this fomrula:

F1 = k*q1*Q

where Q is the charge of the third point (4.5 micrC)

F1 = (9x10^9 Nm^2/C^2)(2.5x10^-6 C)(4.5x10^-6 C)/(0.5^2)
F1 = 0.405 N *(3/5)
F1 = 0.243 N

I then doubled this value to account fro the charge at point 2 as well, thereby giving me an F = 0.486 N

It says this is correct, but I'm a little unsure why I multiple 0.405 by (3/5). Could someone explain this to me?
 
  • #6
Muphrid
834
2
Why don't you find the x and y components of the force separately and see?
 
  • #7
Northbysouth
249
2
Is it equivalent to sin(θ) where θ is equal to the angle between point 3 (4.5 microC) and point 1 (2.5 microC)?

My understanding is that the equation:

F = K11q2/r^2 gives me the force of one charge acting on another. Thus this would mean that the force of charge 1, which I found to be 0.405 N would be acting from charge 1 to charge 3 (point Q with a charge of 4.5 microC)

Thus:

F1x = (0.405N)(0.3/0.5) = 0.243

F1y = (0.405N)(0.4/0.5) = 0.324

But I'm not sure I understand this either. I can take the magnitude of these two values and it'll give me 0.405N, which doesn't surprise me. Can you give me some more hints?
 
  • #8
Muphrid
834
2
When you add the forces from two points, you should add the components separately.

For one of the charges, you should get negative force component for one of the directions.
 

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