Electric Field of x-axis line of charge at the origin.

[fitzfool]

Homework Statement

Calculate the charge and electric field at the origin of non-uniform line of charge on the x-axis from -10cm to 10 cm on the x-axis. Where linear charge density is given by 2x^4

Homework Equations

dq=lambda dx
de=(k dq)/r^2
E=integral of (Kdq)/r^2 = k integral of (lambda dx/r^2)

The Attempt at a Solution

Charge is easy you just integrate the linear charge density equation from -10 to 10.

For the electric field at the origin I believe it is zero but am not sure how to calculate it.

Every example problem puts the point P at some distance away from the line of charge not at the center or even on the line of charge.

The only thing I can think of is that I need to treat the single line of charge like two separate lines(by a extremely small negligible amount) starting at the origin. Then integrate from -10 to 0 and 0 to 10 adding them together to get the total Electric Field.

However, I am not sure if I am even allowed to do it this way.

 

[collinsmark]

Hello fitzfool,

Welcome to Physics Forums!

Homework Statement

Calculate the charge and electric field at the origin of non-uniform line of charge on the x-axis from -10cm to 10 cm on the x-axis. Where linear charge density is given by 2x^4

Homework Equations

dq=lambda dx
de=(k dq)/r^2
E=integral of (Kdq)/r^2 = k integral of (lambda dx/r^2)

The Attempt at a Solution

Charge is easy you just integrate the linear charge density equation from -10 to 10.

For the electric field at the origin I believe it is zero but am not sure how to calculate it.

Every example problem puts the point P at some distance away from the line of charge not at the center or even on the line of charge.

Yes, it doesn't make sense to put the "test charge" on top of an actual charge, since the electric field blows up as r → 0.

But it's okay for this problem, since there technically is no charge at the origin. The charge density, λ = 2x⁴ is 0 when x = 0, so it's okay to put the "test charge" at the origin in this case.

The only thing I can think of is that I need to treat the single line of charge like two separate lines(by a extremely small negligible amount) starting at the origin. Then integrate from -10 to 0 and 0 to 10 adding them together to get the total Electric Field.

However, I am not sure if I am even allowed to do it this way.

Treat the line as an infinite number of infinitesimal point charges. The charge dQ of each infinitesimal point charge is λdx. Integrate Coulombs law for point charges, using that.

[Edit: you even have the proper equation in your list of relevant equations. Just change r to x.]

[Another edit: And yes, you should find that it goes to zero due to symmetry. You'll need to treat the negative x-axis differently than the positive x-axis, because the unit vector \hat {a_r} is in different directions for each. So yes, treating it as two different lines is the right way to go here. ]

[Still, yet another edit: This problem is a good example of why vector calculus is more tricky than plain old run-of-the-mill calculus. Remember, electric field is a vector, that has magnitude and direction. Coulomb's law states, in differential form,

\vec {dE} = k \frac{dQ}{r^2} \hat{a_r}
And it's those pesky unit vectors, \hat{a_r} that cannot be ignored, and can make things tricky. In this problem, \hat{a_r} is different for points on the negative x-asis than it is for points on the positive x-axis. So you'll need to take that into account. This will not be the last time that the pesky unit vectors make things tricky. You can look forward to more of this in future problems. ]

 

[fitzfool]

Hmm, thank you very much for the reply and all of the edits. They were very helpful.

However, I wasn't even thinking of the unit vectors until you posted your last reply.

Taking into consideration the unit vectors I'm getting,

For the left side K ∫2x⁴ / x² from -10 to 0 is K .2/3 x³ from -10 to 0 is -5.9E11 multiply that by cos(0) is -5.9E11
For the right side I get 5.9E11 cos(180) is -5.9E11.
-5.9E11+-5.9E11 is -11.8E11. So I must be messing up a sign somewhere. I'm not sure where. Maybe I could get a hint?

 

[collinsmark]

Hmm, thank you very much for the reply and all of the edits. They were very helpful.

However, I wasn't even thinking of the unit vectors until you posted your last reply.

Taking into consideration the unit vectors I'm getting,

For the left side K ∫2x⁴ / x² from -10 to 0 is K .2/3 x³ from -10 to 0 is -5.9E11 multiply that by cos(0) is -5.9E11
For the right side I get 5.9E11 cos(180) is -5.9E11.
-5.9E11+-5.9E11 is -11.8E11. So I must be messing up a sign somewhere. I'm not sure where. Maybe I could get a hint?

I think the problem was with the evaluation of the negative x-axis part. Remember that (-10)³ is a negative number, and since that limit is on the bottom part of the integral, that also gets evaluated as negative -- a negative times a negative ...

\vec E = k \int_{-10}^0 2 x^2 dx \ + \ -k \int_0^{10} 2x^2 dx
= \frac{2}{3}k \left[0^3 - \left(-10 \right)^3 \right] -\frac{2}{3}k \left[ 10^3 - 0^3 \right]

You should be able to take it from there.

 

[fitzfool]

Thank you very much.
Can't believe I missed that 0 - a negative number is a positive number!