Electric fields and positive point charge

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SUMMARY

The discussion focuses on calculating the electric field generated by a positive point charge q at position x=a and a negative point charge -q at x=-a. The magnitude of the electric field at x=0 is determined using the formula E=1/4πε₀ * |q|/r², resulting in 2q/4πε₀a². For part (b), participants emphasize the necessity of applying the superposition principle to derive an expression for the electric field along the x-axis, specifically between -4a and +4a, by summing the contributions from both charges.

PREREQUISITES
  • Understanding of electric fields and point charges
  • Familiarity with the superposition principle in electrostatics
  • Knowledge of the formula for electric field strength, E=1/4πε₀ * |q|/r²
  • Basic graphing skills for visualizing functions
NEXT STEPS
  • Study the superposition principle in electrostatics
  • Learn how to derive electric field expressions for multiple point charges
  • Explore graphing techniques for electric field functions along the x-axis
  • Investigate the effects of charge position on electric field direction and magnitude
USEFUL FOR

Students and educators in physics, particularly those focusing on electrostatics, as well as anyone interested in understanding electric fields generated by multiple point charges.

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A positive point charge q is placed at x=a and a negative point charge -q is placed at x=-a.
a: find the magnitude of the electric field and the direction at x= 0.
b: derive an expression for the electric field at points on the x-axis. use results to graph the x-component of the electric field as a function of x, for values x between -4a and +4a.

I would get the first part by using e=1/4pi(e0) * modulus q / r^2

so in this case i would get 2q/4pi(e0)a^2 .. right?

now how do i got about doing question b... don't even know where to start.
 
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(a) is right
(b) is asking you to find an expression that gives you the magnitude of the E field for any given x value. Might want to apply the superposition principle here.
 
whozum said:
(a) is right
(b) is asking you to find an expression that gives you the magnitude of the E field for any given x value. Might want to apply the superposition principle here.

in principal would it just be using the same formula as in the first one.. i qould get new values for a because the test charge would "change" positions between -4a and +4a.

so like if electric field x=-4a then i would get kq/(-3a)^2 - kq/(5a)^2 = E1-E2
because the paricle at -3a is negative and attract but the particle at 5a is positive and repells so the net force would be E1 - E2...
??
doing something wrong right, because it's supposed to be e1+e2 right, but hmm how do i get both direction and magnitude of Enet?

edit: if I'm completely wrong please tell me =)
 
Last edited:
The graph is a separate problem. Give me an expression for the E field of a point charge at any point on an axis. If that's trivial, then try to realize that the E field at any point on the x-axis due to 2 charges is the sum of the E fields due to each charge..
 

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