Point Charges, Maginitudes and Electric Fields.

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Homework Help Overview

The problem involves two point charges, one positive and one negative, and their effects on a third point charge. The participants are exploring the calculation of the total electric force exerted on the third charge by the first two charges, focusing on both magnitude and direction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of electric fields and forces, questioning the validity of simply doubling values derived from single charge interactions. There are attempts to break down forces into components and explore vector addition.

Discussion Status

Some participants have provided calculations and expressed uncertainty about specific steps, particularly regarding the use of trigonometric components and the rationale behind certain multiplications. There is an ongoing exploration of how to correctly combine forces from multiple charges.

Contextual Notes

Participants are working within the constraints of homework guidelines, which may limit the information they can share or the methods they can use. There is also a focus on understanding the underlying physics principles rather than simply obtaining a numerical answer.

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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?
 
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Why do you think you can just double the value?
 
Because there's two point charges.
 
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.
 
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?
 
Why don't you find the x and y components of the force separately and see?
 
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?
 
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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