Newtonian Gravity: The Bisector Approximation

[MetaFrizzics]

Suppose we are given a rod with electic charges distributed uniformly from end A to end B. Show that the electric Field at an arbitrary pt C is approximately along the bisector of \angle ACB .

As the phrasing indicates, the 'bisector' is only an approximation of the direction of the force from a charged rod or string of mass-particles. Let's see why:

Lets look at the proposal of the question in a diagram:

 

 

Now lets look at a simple theorem about circles:

 

 

Now lets choose a test-point which is on the rim of the circle of unit radius

around the geometric centre (midpoint) of the rod. This will make the directions from our test-point to each end of the rod a right-angle.

 

First lets take the most simple case, of two equally spaced like-charges on a rod,

or else a string made of two Hydrogen atoms (a diatomic molecule of H gas):

 

 

That is, by simple inspection, since the force from the upper charge is greater than the lower, and the total angle is 90 degrees, the bisector would be 45 degrees, but the resultant force is NOT 45 degrees, since this would only be possible if the forces were equal.

The pull from each end of the rod is covered by the same formula for charge or mass:

 

[math] \vec{F} = \frac{Gmm}{d^2} [/math] or,

 

[math] \vec{F} = \frac{Kqq}{d^2} [/math]

 

We can set the masses or charges = 1, and choose units of distance to make the constants = 1 as well, simplifying both equations to:

 

[math]F = \frac{1}{d^2} [/math] in the direction of each charge (or mass-particle)

 

And the total force upon a test particle will be:

 

[math]\vec{F} = \vec{F_{top}} + \vec{F_{bot}} [/math]

 

by a vector addition of the force from each end of the rod (barbell/molecule).

When we convert this to a single vector, e.g. by conversion from Cartesian form to Polar form, we will find that in general it will *NOT* be a 45 degree ([math]\frac{\pi}{4}[/math]) vector (from either component vector).

 

Now let me sum up:

 

The line through the geometric centre does not bisect the total end-to-end angle.

The line for the direction of force doesn't do this either, because even though the charges are equally spaced, they do not form a continuum, but are discretely located quantized points.

The bisector APPROXIMATION is only valid when the # of charges or mass-points is large enough to be treated as a continuum.

We could also say that it is a good approximation when the distance from the rod is much greater than the length ( d >> L ). But this is just a trivial variation of the Centre of Mass Approximation, since obviously rods and other objects can be treated as point-masses at large distances.