Electric fields and positive point charge

In summary, the conversation discusses finding the magnitude and direction of the electric field at x=0 when a positive point charge q is placed at x=a and a negative point charge -q is placed at x=-a. The formula e=1/4pi(e0) * modulus q / r^2 is used to calculate the magnitude of the electric field, which is 2q/4pi(e0)a^2. The discussion then moves on to deriving an expression for the electric field at points on the x-axis and using the superposition principle to calculate the net electric field. The conversation also mentions creating a graph of the x-component of the electric field as a function of x for values between -4a and +4a
  • #1
Pixter
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.
 
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  • #2
(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.
 
  • #3
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:
  • #4
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..
 

1. What is an electric field?

An electric field is a region around a charged object or point charge in which other charged particles experience a force. It is created by the presence of a charged object and can be either positive or negative.

2. How is an electric field affected by a positive point charge?

A positive point charge creates an electric field that radiates outward in all directions. The strength of the electric field decreases as the distance from the point charge increases.

3. What is the direction of the electric field created by a positive point charge?

The electric field created by a positive point charge always points away from the charge. This is known as the direction of the electric field lines.

4. How is the electric field strength affected by the distance from a positive point charge?

The strength of the electric field decreases as the distance from the positive point charge increases. This relationship is known as the inverse-square law, where the electric field strength is inversely proportional to the square of the distance from the point charge.

5. How can the electric field around a positive point charge be calculated?

The electric field around a positive point charge can be calculated using the equation E = kQ/r^2, where E is the electric field strength, k is the Coulomb's constant, Q is the magnitude of the charge, and r is the distance from the charge. This equation only applies to point charges, and for larger charged objects, the electric field must be calculated using calculus.

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