Electric field due to a finite line of charge

  • #1

Homework Statement


A rod of length 25 cm has a uniform linear charge density of 7 μC/m. Determine the Electric Field at a point P located at a perpendicular distance 69 cm along a line of symmetry of the rod.


Homework Equations


E=k*charge density(integral(dx/(d^2+x^2)^3/2)


The Attempt at a Solution


x=.69tan(theta)
dx=(.69)sec^2(theta)dtheata
Thus,
.125=pi/18
Hence, I have found
=(9*10^9)*(7*10^-6)*2.1004*(sin(pi/18)-sin(-pi/18))
My final answer is 45956.1 but it's wrong.

Thank you for any input at all. Also, I'm not sure what mode is my calculator suppose to be on.
 

Answers and Replies

  • #2
kuruman
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What is the meaning of

.125 = pi/18?

If that equation is correct, then pi = 18*.125 = 2.25, not true.

To find the sine at the extreme angles, you don't need to calculate any angle. Just use the basic definition

sine = opposite/hypotenuse.
 
  • #3
5
0
I am in a course currently studying this topic, and this is how I understand and solved the problem:

Electric field due to a continuous charge distribution:
E (as a vector) = [tex]\int[/tex]kdq / r^2 (in r vector direction)
So, dE = kdq / r^2 = k[tex]\lambda[/tex]dx / r^2

The E field has both x and y component, but if you draw the diagram as I see it, there is point p in the middle and directly above the uniformly charged line (ie along a line of symmetry). In that case, the x component will be 0 and there will just be the y.

Also, since this point is directly in the middle it cuts the charged line half, and so we let it's length (L) be L/2 until the intersection point and L/2 past the intersection point. This makes two right triangles with [tex]\Theta[/tex](1) representing the top angle for the first, and [tex]\Theta[/tex](2) representing the top angle for the second. It is difficult to explain this without a drawing...

So, we just need to find the y component of the E field:

E(y component) = dEcos[tex]\Theta[/tex] = (k[tex]\lambda[/tex]dx / r^2) (y / r) = k[tex]\lambda[/tex]ydx/r^3

You can use trig substitution to solve this integral or use some relationships in the graph to simplify it. Notice cos[tex]\Theta[/tex](2) = y / r. So, 1/r = cos[tex]\Theta[/tex](2) / y.
Also, tan[tex]\Theta[/tex] = x / y. So, x = ytan[tex]\Theta[/tex] and dx = ysec^2[tex]\Theta[/tex] d[tex]\Theta[/tex].

Plugging this stuff in we get dE(y) = k[tex]\lambda[/tex]yysec^2[tex]\Theta[/tex]d[tex]\Theta[/tex]cos^3[tex]\Theta[/tex](2) / y^3

Simplifying should get k[tex]\lambda[/tex]cos[tex]\Theta[/tex]d[tex]\Theta[/tex] / y

E(y) = k[tex]\lambda[/tex] / y [tex]\int[/tex]cos[tex]\Theta[/tex]d[tex]\Theta[/tex]

Solving this integral gives k[tex]\lambda[/tex] / y (sin[tex]\Theta[/tex](2) - sin[tex]\Theta[/tex](1))

Since [tex]\Theta[/tex](2) = -[tex]\Theta[/tex](1) in the graph, the sines can be written as (sin[tex]\Theta[/tex] - sin(-[tex]\Theta[/tex])) = 2sin[tex]\Theta[/tex]

Thus, E(y) = (2k[tex]\lambda[/tex] / y) sin[tex]\Theta[/tex]

Notice from the graph that sin[tex]\Theta[/tex] = x/r = (1/2L)/([tex]\sqrt{(1/2L)^2 + y^2}[/tex]
So, E(y) = (2k[tex]\lambda[/tex]/y) ((1/2L)/([tex]\sqrt{(1/2L)^2 + y^2}[/tex])

The expression [tex]\lambda[/tex]L can be rewritten as Q, since it is the charge per unit length. So, on top, the 2 and (1/2) cancel leaving you with just kQ in the numerator.

Plugging in the numbers gives (8.99x10^9)(7x10^-6) / ((.69) * ([tex]\sqrt{1/2(.25)^2 + (.69)^2}[/tex]))
The resulting calculation is 128042.6956 N/C
 
Last edited:
  • #4
I don't understand the difference between theta 1 and theta 2. Could you please explain to me on the graph?
 
  • #5
I get it. Thank you for your help.
 
  • #6
5
0
I get it. Thank you for your help.

Ok, if you got it then that's good. Yeah, I was thinking the explanation doesn't really do too much good without being able to visualize it...
 

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