Exploring Electric Fields from a Charge Distribution

[Mdhiggenz]

1. Homework Statement [/b]

Positive charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point

charge q is located on the positive x-axis at x=a+r , a distance to the right of the end of Q.

Observe the figure below

My goal is to try to explain the problem to make sure I understand it since a similar one will be on my exam, and also get some questions that I don't understand addressed. (:

1. Calculate the x-component of the electric field produced by the charge distribution Q at points on the positive x-axis where x>a.

$\lambda$= $\stackrel{Q}{a}$

dQ=$\-lambda$ dx (I don't know why lambda is negative)

dq=-$\stackrel{Q}{A}$

First things First I know that the distance of Q is a and the distance from Q to a test point q+ is = a+r which the problem statement sets to x for simplicity.

-$\aintx$ $\stackrel{Q}{a}$dx

Here is where I get confused I don't understand why the insert $\stackrel{k}{x^2}$
not including the point charge q?

after that it is just simple integrals and algebra.

Ultimately getting an answer of $\stackrel{Qk}{a}$($\stackrel{1}{a}$-$\stackrel{1}{a+r}$)

2. Calculate the magnitude of the force that the charge distribution that Q exerts on q.

Here I'm also confused isn't the answer just a simplified version of part 1?

$\stackrel{kQa}{r(a+r)}$

Thanks for the help!

 

[jbsteele]

2. Homework EquationsE = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^2}F = qE3. The Attempt at a Solution1. Calculate the x-component of the electric field produced by the charge distribution Q at points on the positive x-axis where x>a.\lambda= \stackrel{Q}{a}dQ=\-lambda dx (I don't know why lambda is negative)dq=-\stackrel{Q}{A}The electric field at any point along the x-axis due to the charge distribution Q is given by:E_{x} = \frac{1}{4 \pi \epsilon_{0}} \int_{0}^{a} \frac{\lambda}{(x-a)^2} dxWhere \lambda is the linear charge density, i.e. the charge per unit length of the line charge.Therefore, the x-component of the electric field produced by the charge distribution Q at points on the positive x-axis where x>a is:E_{x} = \frac{Q}{4 \pi \epsilon_{0}} \left(\frac{1}{a} - \frac{1}{a+r}\right)2. Calculate the magnitude of the force that the charge distribution that Q exerts on q.The magnitude of the force exerted on q by the charge distribution Q is given by:F = qEWhere q is the charge on q and E is the electric field at the position of q due to the charge distribution Q.Therefore, the magnitude of the force that the charge distribution Q exerts on q is:F = q \frac{Q}{4 \pi \epsilon_{0}} \left(\frac{1}{a} - \frac{1}{a+r}\right)