Electric field of a line charge and point charge

1. May 23, 2007

GravityGirl

A thin rod 39.2 cm long is charged uniformly with a positive charge density of 72.0 mC/m. The rod is placed along the y-axis and is centered at the origin. A charge of +46.5 mC is placed 68.4 cm from the midpoint of the rod on the positive x-axis. Calculate the electric field at a point on the x-axis, which is halfway between the point charge and the center of the rod. (Express your answer in terms of the unit vector x. For example, if the electric field is -10.5x N/C, then input -10.5 N/C.)

ok..so i drew a picture and i see that the E field of te line charge will be in the x direction becuase that i perpendicular to the line

i also see that the point charge will be in the x direction as well becuase it is located on the x-axis

so since both charges are postive and the point where i want to know the E field at is between them i am going to subtract their individual E fields at that point to get the net

E=Eline-Epoint

Eline=2K(lamda)/r where r= half the distance betwen the line and point charge

Epoint=kq/r^2

so i have E=(2k(lamda)/r)-(kq/r^2)

does this look right?....if not, please tell me what i am doing wrong

2. May 23, 2007

Staff: Mentor

OK.

The field from that charge will point along the negative x-axis.

OK. But I would prefer that you viewed it as adding the positive field from the line charge to the negative field from the point charge.

That formula applies to an infinitely long line charge. For a finite line charge you need to integrate.

OK.

3. May 23, 2007

neutrino

Oops. I must've misread what GravityGirl wrote.

4. May 23, 2007

GravityGirl

ok so if i am to integrat across the line charge then i would have

integral of Kdq/r^2=kQ/x(x-L) where L= the lenght of the rod and x= the distance to the point where i want to find the E field

so now i have (KQ/x(x-L))-(kQpoint/r^2)=E

but doesn't gauss's law also work (what i did above)

5. May 23, 2007

Staff: Mentor

How did you get that answer? Doesn't look right to me. (What if x = L?)
Gauss's law is only useful when there is enough symmetry to simplify the calculation of the electric flux. That's true with an infinite line of charge, since you know the field must be radial; but not true with a finite line of charge.

6. May 23, 2007

GravityGirl

i got the simplified answer to the integral from my book....but what about if is had ....

=k/r^2(intergral of dq)=kQ/r^2

does that look better??

7. May 23, 2007

Staff: Mentor

That's only true for a point charge at one location. In a line charge, each charge element is at a different distance and direction. Read this: Electric Field of Line Charge

8. May 23, 2007

GravityGirl

i see that the equation Ez is the same equation that i wa using in the firs attempt

9. May 23, 2007

GravityGirl

so should Eline=2k(lamda)/r?

10. May 23, 2007

Staff: Mentor

No it isn't. Make sure you are looking at the right equation.

No, as I said before, that only applies to an infinitely long line of charge.

11. Feb 14, 2010

sonutulsiani

Ok here can't we just take it as a distance of 68.4/2 = 34.1 cm from the midpoint of the rod? And then just integrate k dq/r^2? with r = 34.1 cm which can be converted to metres later.

12. Feb 14, 2010

Staff: Mentor

You cannot treat the line charge as if its charge were concentrated at its center, if that's what you're thinking.

13. Feb 14, 2010

sonutulsiani

No I am not thinking that

I am thinking that if we take the symmetry from the line charge from both the sides, then we get the x component of the E field at 34.1 cm right?

14. Feb 15, 2010

Staff: Mentor

I still don't understand what you mean. To calculate the field from that line charge you must integrate the field from each element of charge along the line. By symmetry, you know that there will only be an x component of the field.

15. Mar 1, 2010

Dosmascerveza

you can integrate the function dE=kdQ/(r^2). dQ=lambda dy where lambda is the linear charge density Q/a where a is the length of the line of charge. and you find the y component by multiplying by the unit vector sin(phi) where phi is the angle between the x axis and r where r is the distance from P(x,y) to the upper limit of your line of charge. r=sqrt(x^2+y^2)

SO YOU GET...

S dE_y= kQ/(a) S[(y)r^(-3/2)]dy.

What you will get if you integrate the right side from 0 to a is the y component of your Electric field. To find the x component uses a similar process but the integral is a bit trickier (trig substitution) Then you can plug and chug away.

there is probably an easier way to do this but i think the long way helps build an intuition because now its clear as day for me.