# 2 solutions to the same problem but which is the correct one?

1. Sep 23, 2004

### stunner5000pt

DO NOT USE GAUSS LAW BECAUSE I HAVEN'T LEARNT IT
You are given a positively charged rod of length L and of univform charge Q. TYhere is a positive test charge qo placed alongside this rod like the figure

+++++++++++++++++++++++++++++++++<-------distance x------>Q

find the net force on Q

First of all there is no y or z component so F refers to the x axis only

One solution :

Let dq = YdL where Y is charge density
then dF = k dq Q/(L+x)^2

dF = kqYdL/(L+x)^2

integrating you get

F = kqY dL/L+x)^2 from 0 to L

then F = kqYL/x(L+x)

then resub for Y = Q /L

F = KqQ/x(x+L)

is there anything wrong with this solutioin please point out a mistake i don't seem to ppick up on.

My prof suggested this way

Divide the rod like so

++++++++++++++++<------distance x--------->Q
<----L/2---><---------(x - L/2)------------------->
then dq = Y dL

dF = k dq Q / (x-L/2)^2

dF = kYdLQ/ (x-L/2)^2

F = kYQ (integrate from 0 to L) dL / (x-L/2)^2

solving gives kqQ/x(2x-L) why this inconsistency of solutions i know one is wrong, but which one??
i have a feeling that the integration for the second one is not done properly...

Last edited: Sep 23, 2004
2. Sep 23, 2004

### Staff: Mentor

Please describe where q0 is placed. For example, if the charged rod extends from x = 0 to x = L, where is q0?

3. Sep 23, 2004

### stunner5000pt

it i sm y typing mistake qo = Q

and Q is a distance x from the tip of the rod

4. Sep 23, 2004

### Staff: Mentor

Your notation is a bit confusing, but your answer is correct. I believe you probably solved it correctly. I'd write it like this:
Let an element of charge on the rod be: $dQ = \lambda dx$, where I use "x" as the variable specifying the distance of the charge element from $q_0$. (I'll call the distance of the test charge from the end of the rod $x_0$.)

So $dF = k \lambda q_0/x^2 dx$.
Integrate from x = $x_0$ to $x_0 + L$.
The answer is identical to yours.

I have no idea what your prof was telling you to do.

Last edited: Sep 24, 2004
5. Sep 24, 2004

### illusion88

The method is correct but something is wrong.
We take an element 'dl' at a distance 'l' and not 'L'.then apply the same formulae and integrate from x to x+L.this gives u the correct result.

sriram