• Support PF! Buy your school textbooks, materials and every day products Here!

Electric field and force

  • Thread starter Nasserz
  • Start date
4
0
[SOLVED] Electric field and force ...

1. Homework Statement
We have 2 identical thin rods , each has a length of 2a and carry a charge of +Q , uniformly distributed along their lengths , and both lie on the horizontal X-axis , and the distance between their centers is "b" ...
I need to calculate the electric field and the magnitude of the force exerted by the left rod on the right one.


2. Homework Equations
I know that F=Q*E
and the electric field done by a charged rod on a point in its axis is E= K*Q/d(d+L) where L is the length of the rod , and d is the distance between the rod and the point.


3. The Attempt at a Solution
Well I didn't know how to start :S , no need for a full solution , I just need a push and tell me how to start to solve it .. thanks.
 

Answers and Replies

Shooting Star
Homework Helper
1,975
4
You have to integrate along the length of the right rod.

If [itex]\lambda[/itex] is the linear charge density of the right rod, then find the force due to left rod on an element of length dx on the right rod which is at a distance of x from the left rod. Now integrate, putting the proper limits of x, that is, the values of x on the left and right extremities of the rod on the right respectively.
 
4
0
can someone explain a litle bit more .. didn't really understand ... my english isnt that good
 
Shooting Star
Homework Helper
1,975
4
2. Homework Equations
I know that F=Q*E
and the electric field done by a charged rod on a point in its axis is E= K*Q/d(d+L) where L is the length of the rod , and d is the distance between the rod and the point.


3. The Attempt at a Solution
Well I didn't know how to start :S , no need for a full solution , I just need a push and tell me how to start to solve it .. thanks.
So you don’t need a push, but a slightly bigger shove. Here goes!

Draw a diagram first of the two rods. Take a small length dx on the right rod, at a distance x from the right tip of the left rod. This dx length has got a charge of dq = [itex]
\lambda[/itex]dx.

What is the field at this point due to the charge of the left rod? This is at a distance of x from the right end of the left rod. Apply the formula you have written. (I have not checked it, but looks all right.) The field should be then KQ/[x(x+2a)].

So the force on this dq on the right rod should be given by:

dF = Field*Charge = KQdq/[x(x+a)] = K[itex]\lambda[/itex]dx/[x(x+a)] => the total force F should be given by:

[tex]F = \int^{b}_{b-2a}\frac{KQ\lambda dx}{x(x+2a)}[/tex].
 
Last edited:
4
0
thanks mate , thats very helpful ... :)
 
4
0
I continued the integration .. and I got as final answer : F = (KQ^2/4a^2)ln(b^2/b^2 - 4a^2)
I think it is right , can you confirm this ?
thanks alot mate , I really appreciate your help ^_^ , you are the best.
 
Shooting Star
Homework Helper
1,975
4
I continued the integration .. and I got as final answer : F = (KQ^2/4a^2)ln(b^2/b^2 - 4a^2)
Bracket is missing: (KQ^2/4a^2)ln(b^2/(b^2 - 4a^2)).
 

Related Threads for: Electric field and force

  • Last Post
Replies
13
Views
9K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
17K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
665
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
8
Views
2K
Top