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In lecture 2 of http://oyc.yale.edu/physics/physics/fundamentals-of-physics-ii/content/class-sessions [Broken], around the 50 minute mark, the professor calculates the value of an electric field at a point (

This electric field is due to two equal point charges: one positive, one negative. The positive charge is positioned on the

[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

I began like this, without making any assumptions as to where on the

[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) 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.
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