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Electric fields and positive point charge

  • Thread starter Pixter
  • Start date
30
0
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.
 
2,208
1
(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.
 
30
0
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 completly wrong plz tell me =)
 
Last edited:
2,208
1
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 thats 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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