Electric Field distributed Uniformly along the X-axis

[shadowfox745]

Homework Statement

A charge Q is distributed uniformly along the x-axis from x1 to x2. If x1=0m, x2= 4.82m, and the charge Q = 3.13 microcoulombs, what is the Magnitude of E of the electric field at x0= 11.7m on the x-axis. Assume that x0>x2>x1. Answer in unitsof N/C.

Homework Equations

dE= kdq/r^2
K=8.99x10^9

The Attempt at a Solution

Ok so I got the integral down as (KQ)/((x2-x1)(x0-x)^2) from x1 to x2. And I simplified the integration KQ/(x2-x1) times the integral of 1/(x0-x) from x1 to x2. I then plugged in the numbers and got 3.099x10^3 N/C which is incorrect. Did I miss any crucial steps?

 

[anathema]

Hello,

Your approach and equations seem to be correct. However, make sure you are using the correct values for x1, x2, and x0 in your calculations. The given values for x1 and x2 are in meters, but the value for x0 is not specified. It could be a typo, so make sure to double check the values you are using. Also, make sure to convert all units to their SI units before plugging them into the equation.

Another thing to consider is the direction of the electric field. Since the charge is distributed along the x-axis, the electric field will be in the same direction at all points on the x-axis. Therefore, the magnitude of the electric field will be the same at x0 as it is at x1 or x2.

I hope this helps!

 

[baba89]

Your approach is correct, however, there seems to be a mistake in your integration. The correct integration should be (KQ)/((x2-x1)(x0-x)) from x1 to x2. This will give you a value of 3.099x10^3 N/C, which is the correct answer. Double check your integration and make sure to use the correct limits of integration.