Physics help more of a proof type.?

Question

http://i328.photobucket.com/albums/l341/CristyAustin/Scan0002-1.gif

I dont even know where to start.

Answer 1

This is not going to be an answer, but in reading the problem, I sense a typo in the question.

Point A is not on the perpendicular bisector of the rod.  Point B

IS on the perpendicular bisector.

Any rod that is charged exerts force on a point particle not on the rod, due to axial and radial components.

If indeed the question is intended to be about point A as labeled, then the radial force due to radial component is neglected.  The entire force is due to axial force.

Someone else already solved this, but the explanation is a bit tricky to understand.  I had to read over it several times!  It is set up for three dimensions, but your situation is simpler in that it is only two dimensions, and one of the dimensions is neglected, so it essentially becomes a one dimensional problem. Think of the rod as small parts, each having an influence on the point particle. You will need to determine the force acting on the point due to one small chunk of the rod...then the integral is formed when you account for the sum of the forces due to each chunk of rod...

Now, if the question is truly intended to focus on point B...then you will need the radial force, but the axial force is neglected.

If your point is not on the axial line or the perpendicular bisector, you will need to calculate both components.

Hope this helps!  SP

"The result depends on only two variables: the radial distance between the

point and the rod, and the axial distance. Let's take a reference frame X,

Y, Z such that the rod is along the Z axis and centred at the origin, thus

starting at z = -l / 2 and ending at z = l / 2. If x, y, z are the

coordinates of the point, the radial and axial distance are respectively

r = sqrt(x^2 + y^2) and h = z.

Each infinitesimal element of the rod gives a force which radial and axial

components are:

dF_r = k r dz / sqrt((h-z)^2 + r^2) / ((h-z)^2 + r^2)

dF_a = k (h-z) dz / sqrt((h-z)^2 + r^2) / ((h-z)^2 + r^2)

Finally the total force is their integral over z between -l / 2 and l / 2

F_r = k r Int dz / ((z-h)^2 + r^2)^3/2

F_a = k Int (h-z) dz / ((z-h)^2 + r^2)^3/2

Combining them, the force is:

F = sqrt(F_r^2 + F_a^2)

Good luck.

Your may also consider an infinite rod. It is then much simpler since the

axial component is zero and the force is proportional to 1 / r."

Answer 2

Seems like you're a bit over your head eh? (Just answered your last one lol). Anyway, for this, you want to use the idea that potential can be expressed due to a point charge as q/(4pi epsilon *r). So since it's a continuous distribution, you want to integrate all the "point charges" along the length of the rod. So set up the integral.

You are integrating little bits of charge, let's call them dq. Now you need to make a substitution to get the integral in terms of a length variable. So you have a charge distribution lamda. The amount of charge per tiny unit of length, since lamda is charge/length, is lamda * dx.

So make the necessary substitutions.

lamda/(4*pi*epsilon) * int(0,L)[ dx/(x+d)]

Factoring out the constants and having the distance from the point to the rod as x+d or x-(-d) since if it's say at x = 0, the distance is 0 + d, etc.

And since it just wants the integral, that's it. If you wanted to do the integral it would be a natural log expression evaluated between the limits.