- #1
Rasalhague
- 1,383
- 2
In lecture 2 of http://oyc.yale.edu/physics/physics/fundamentals-of-physics-ii/content/class-sessions , around the 50 minute mark, the professor calculates the value of an electric field at a point (x,0,0) on the x-axis of a cartesian coordinate system. (The video can be rather slow loading, so hopefully my description will be clear enough to stand alone.)
This electric field is due to two equal point charges: one positive, one negative. The positive charge is positioned on the x-axis at (a,0,0), where a > 0, and there's an equal negative charge at -a. He begins by giving the sum of the fields generated by each charge as
[tex]\frac{q \mathbf{i}}{4\pi\epsilon_0}\left ( \frac{1}{(x-a)^2}-\frac{1}{(x+a)^2} \right ).[/tex]
Am I right in thinking that this formula is only valid when x > a? In that case, the first term will be positive and the second negative, as I think they should be. But if x < -a, then, E-(x,0,0), the component of the electric field vector at (x,0,0) which is due only to the negative charge should point in the positive x direction (attracting a positive test charge), while E+(x,0,0), the component due only to the positive charge will point in the negative x direction (repelling a positive test charge), shouldn't it? And when (x,0,0) is between the two charges, I think, both components will point in the negative x direction.
I began like this, without making any assumptions as to where on the x-axis (x,0,0) is:
[tex]\frac{q \mathbf{i}}{4\pi\epsilon_0}\left ( \frac{x-a}{\left | x-a \right |^3}-\frac{x+a}{\left | x+a \right |^3} \right ).[/tex]
His formula simplifies to
[tex]\frac{qax\mathbf{i}}{\pi\epsilon_0(x^2-a^2)^2}.[/tex]
I guess if a similar simplification is possible in the more general case, the three special cases of where (x,0,0) can be in relation to (a,0,0) and (-a,0,0) need to be checked.
This electric field is due to two equal point charges: one positive, one negative. The positive charge is positioned on the x-axis at (a,0,0), where a > 0, and there's an equal negative charge at -a. He begins by giving the sum of the fields generated by each charge as
[tex]\frac{q \mathbf{i}}{4\pi\epsilon_0}\left ( \frac{1}{(x-a)^2}-\frac{1}{(x+a)^2} \right ).[/tex]
Am I right in thinking that this formula is only valid when x > a? In that case, the first term will be positive and the second negative, as I think they should be. But if x < -a, then, E-(x,0,0), the component of the electric field vector at (x,0,0) which is due only to the negative charge should point in the positive x direction (attracting a positive test charge), while E+(x,0,0), the component due only to the positive charge will point in the negative x direction (repelling a positive test charge), shouldn't it? And when (x,0,0) is between the two charges, I think, both components will point in the negative x direction.
I began like this, without making any assumptions as to where on the x-axis (x,0,0) is:
[tex]\frac{q \mathbf{i}}{4\pi\epsilon_0}\left ( \frac{x-a}{\left | x-a \right |^3}-\frac{x+a}{\left | x+a \right |^3} \right ).[/tex]
His formula simplifies to
[tex]\frac{qax\mathbf{i}}{\pi\epsilon_0(x^2-a^2)^2}.[/tex]
I guess if a similar simplification is possible in the more general case, the three special cases of where (x,0,0) can be in relation to (a,0,0) and (-a,0,0) need to be checked.
Last edited by a moderator: