Point Charges, Maginitudes and Electric Fields.

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SUMMARY

The discussion focuses on calculating the total electric force exerted on a third point charge (Q = 4.5 microC) by two point charges (2.5 microC and -2.5 microC) positioned at specific coordinates. The participant initially miscalculated the electric field by incorrectly doubling the force from one charge instead of considering vector components. The correct approach involves calculating the forces from each charge separately and then combining their vector components to find the total force acting on charge Q. The final force calculated is 0.486 N, which accounts for the direction and magnitude of forces from both charges.

PREREQUISITES
  • Understanding of Coulomb's Law and electric force calculations
  • Familiarity with vector addition and components
  • Knowledge of the Pythagorean theorem for distance calculations
  • Basic principles of electric fields and forces
NEXT STEPS
  • Study vector addition in physics to understand force components
  • Learn about electric field calculations using E = kq/r^2
  • Explore the concept of superposition in electric forces
  • Review trigonometric functions in relation to force direction and angles
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Students studying electromagnetism, physics educators, and anyone involved in solving problems related to electric forces and fields.

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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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