Solving for Electric Field of Uniform Line Charge | Coulomb's Law Example

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Homework Statement

A uniform line charge extends from x = - 2.7 cm to x = + 2.7 cm and has a linear charge density of lambda = 6.5 nC/m.

(a) Find the total charge.

Find the electric field on the y-axis at the following distances.
(b) y = 4 cm



2. The attempt at a solution

It was easy to find the charge...

Q = 6.5 nC/m * 0.054 m = 3.51nC*10e-1

..I'm having issues with how to approach the second part, I don't know whether to use an equation, I'm not even sure which equation to use, I know I need to use Coulomb's Law but I don't know how to use it. Thanks for the help in advance!

 

[chrisk]

Use

$$E = \int\mbox{dE}$$

where

$$dE = K\frac{\lambda\mbox{dx}}{r^2}cos\theta$$

If you draw the configuration the distance r is from a charge element to a point on the y-axis thus forming a right triangle. The horizontal components of E will cancel (equal and opposite) so only the vertical component of E from each charge element will contribute. Express cosine theta in terms of x.

 

[G01]

You have to use calculus and coloumb's law to solve for the field of a finite line charge.

HINT: Find the field from an infinitesimal point on the line using Coloumbs law, then integrate over the length of the line.

 

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Okay, so I tried integrating, I don't think I did it right though. I don't know exactly how to express cos(theta) in terms of x but this is what I tried, please correct me if I'm wrong...

r = 0.0483
theta = 55.981

E = integration(8.99*10^9 * (6.5*10^-9)/(0.0483)^2 *cos(55.981)) with respect to x from -0.027 to 0.027

 

[chrisk]

I erred with the component of E. Since the horizontal components cancel, the vertical component is dEsin(theta). So, the integral is

$$\int\mbox{dE} = \int\mbox{K}\frac{\lambda\mbox{dx}}{r^2}sin\theta$$

Now,

$$r^2=x^2+y^2$$

and

$$\mbox{sin}\theta=\frac{y}{\sqrt{x^2+y^2}}$$

where y is constant (the fixed height to evaluate E). Now, you have a single variable integral in x so set the limits for x and this will give you the desired result. Or you could set the limits from one end of the line to zero then double the result because of symmetry. Use trigonometric substitution to solve the integral.