# Electric field due to a finite line of charge

1. Sep 9, 2009

### Liketothink

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. Sep 10, 2009

### 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.

3. Sep 12, 2009

### 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

Last edited: Sep 12, 2009
4. Sep 13, 2009

### Liketothink

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

5. Sep 13, 2009

### Liketothink

I get it. Thank you for your help.

6. Sep 14, 2009

### aaron2

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...