Force due to uniformly charged rod

[stunner5000pt]

Ok really stuggling with this, and this is the last in the set

lets say there's a rod on the Y and there is a test charge Q located a distance D from the rod on the 45degree line that bisects the x and y axes. (that is the line f(x) = x, identity function) Write an equation in vector component form (?) for hte force on this charge.

Started off like this
I'd do the Forces of x direction i nthree parts for the top , middle and bottom

for the top integrate from 0 to d-L/2 for kQ(lambda)d dL/(2d^2 + L^2/4 - 2dL)

am i on the right track? or am i going off on a tangent

Please do not suggest gauss law i don't know how to use it

 

[vsage]

Gauss' law isn't relevant here IMO. I think you're not telling us all the information here such as the length of the rod or if it is infinite. Could you be more specific? I have a hunch that the problem is worded in such a way that symmetry would help greatly, though

 

[M98Ranger]

The force due to a uniformly charged rod can be calculated using Coulomb's Law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the rod can be considered as a continuous line of charge, with a charge density of λ (lambda) per unit length.

To find the force on the test charge Q located at a distance D from the rod, we can divide the rod into small segments and calculate the force from each segment. This can be done using vector components, as suggested in the prompt.

For the x-direction, we can divide the rod into three parts as mentioned - top, middle, and bottom. The force from each part can be calculated using the equation:

dFx = kQλdL cosθ / (D^2 + L^2/4 - 2DL cosθ)^3/2

Here, k is the Coulomb's constant, Q is the charge of the test charge, λ is the charge density of the rod, dL is the length of the small segment, D is the distance between the test charge and the rod, and θ is the angle between the segment and the line connecting the segment to the test charge.

For the top part, θ = 45 degrees, dL = d-L/2, and the distance between the segment and the test charge can be calculated using the Pythagorean theorem as √(D^2 + L^2/4 - 2DL cos45). Similarly, for the middle and bottom parts, θ = 0 and 90 degrees respectively, and the distances can be calculated using the same formula.

To find the total force in the x-direction, we can integrate the forces from each part over the entire length of the rod. The same process can be repeated for the y-direction, with the only difference being the angle θ.

In summary, you are on the right track by dividing the rod into parts and using vector components to calculate the force. Just make sure to use the correct angle and distance for each part and integrate the forces over the entire length of the rod. I hope this helps you understand the concept better.