Electric field due to a finite line of charge

[Liketothink]

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.

 

[kuruman]

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.

 

[aaron2]

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) = $$\int$$kdq / r^2 (in r vector direction)
So, dE = kdq / r^2 = k$$\lambda$$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 $$\Theta$$(1) representing the top angle for the first, and $$\Theta$$(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$$\Theta$$ = (k$$\lambda$$dx / r^2) (y / r) = k$$\lambda$$ydx/r^3

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

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

Simplifying should get k$$\lambda$$cos$$\Theta$$d$$\Theta$$ / y

E(y) = k$$\lambda$$ / y $$\int$$cos$$\Theta$$d$$\Theta$$

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

Since $$\Theta$$(2) = -$$\Theta$$(1) in the graph, the sines can be written as (sin$$\Theta$$ - sin(-$$\Theta$$)) = 2sin$$\Theta$$

Thus, E(y) = (2k$$\lambda$$ / y) sin$$\Theta$$

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

The expression $$\lambda$$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) * ($$\sqrt{1/2(.25)^2 + (.69)^2}$$))
The resulting calculation is 128042.6956 N/C

 

[Liketothink]

I don't understand the difference between theta 1 and theta 2. Could you please explain to me on the graph?

 

[Liketothink]

I get it. Thank you for your help.

 

[aaron2]

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...