How Does Charge Distribution Affect Electric Field Intensity at the Origin?

[SilverGirl]

Homework Statement

A rod of length l with a uniform charge per unit length lambda is placed a distance d from the origin along the x axis. A similar rod with the same charge is placed along the y axis. Determine the net electric field intensity at the origin.

Homework Equations

Electric Field Intensity is the same as electric field, right? So you would use E=kq/r^2 (substituting d in for r?)

lambda = Q/l

The Attempt at a Solution

I am under the impression the net electric field will be zero, as the charge on the x-axis should cancel out the charge on the y axis, right? Or since the rods are d away from the origin, and not actually on the origin, does this automatically mean the field is zero?

 

[Doc Al]

Homework Equations

Electric Field Intensity is the same as electric field, right? So you would use E=kq/r^2 (substituting d in for r?)

That equation is for the field from a point charge, not a line of charge. Hint: Break the line into segments and add up each of their contributions to the field. You'll have to integrate.

I am under the impression the net electric field will be zero, as the charge on the x-axis should cancel out the charge on the y axis, right?

No. If the line charges where on opposite sides of the origin, then you could conclude that their fields would cancel out at the origin. But that's not the case here.

You'll need to find the field contribution from each line charge and then add them up. Don't forget that electric field is a vector.

 

[SilverGirl]

The fields on the x and y are the same though right? So I find one, and can multiply it by 2?

 

[SilverGirl]

That equation is for the field from a point charge, not a line of charge. Hint: Break the line into segments and add up each of their contributions to the field. You'll have to integrate.

No. If the line charges where on opposite sides of the origin, then you could conclude that their fields would cancel out at the origin. But that's not the case here.

You'll need to find the field contribution from each line charge and then add them up. Don't forget that electric field is a vector.

So dE = kdq/r^2 ?? r^2 being l^2 + d^2 ?

So dE=(klambda*dl)/(l^2 + d^2)

 

[Doc Al]

The fields on the x and y are the same though right? So I find one, and can multiply it by 2?

The magnitudes are the same, but they have different directions. (So you can't just multiply by 2 to get their total.)

 

[SilverGirl]

So the components for the x and y-axis are the same in magnitude, but opposite direction? So once you get the components for the one rod, you are able to just change the direction to find the other?

 

[SilverGirl]

I feel so confused. Right now I have for the rod on the y axis:

Ex = k*lambda*x*(1/x^2)(sinA)(limit A1-A2)

Ex = (k*lambda)/x *{[(l+d)/SQRT(l^2+d^2)] - -(l+d)/SQRT(l^2+d^2)]
Ex = [2k*lambda*(l+d)]/[x * SQRT(l^2+d^2]

 

[Doc Al]

A rod of length l with a uniform charge per unit length lambda is placed a distance d from the origin along the x axis.

Can we assume that this rod is parallel to the x-axis (actually, directly along the x-axis), not perpendicular to it?

 

[SilverGirl]

Can we assume that this rod is parallel to the x-axis (actually, directly along the x-axis), not perpendicular to it?

The picture which I did not include (sorry about that), shows both rods directly on their respective axis

 

[Doc Al]

The picture which I did not include (sorry about that), shows both rods directly on their respective axis

Good, as I had imagined it. The differential field element from a segment dx, ignoring signs, would be given by:

$$dE = \frac{k \lambda}{x^2} dx$$

Take the integral over the proper limits, from x = d to d + l.

 

[SilverGirl]

Thanks very much. I have never done intergration before, so I am not really sure what it means and how to do it. Do you take out the k*lambda/x^2 first?

 

[Doc Al]

Do you take out the k*lambda/x^2 first?

The k*lambda is a constant, so it just tags along. What's the integral (anti-derivative) of 1/x^2?

 

[SilverGirl]

The k*lambda is a constant, so it just tags along. What's the integral (anti-derivative) of 1/x^2?

I have done derivatives before, but never antiderivatives before and am not sure. Just giving a wild guess would be, x ?

 

[Defennder]

An anti-derivative of a function f(x) is simply a function g(x) such that when it is differentiated, it gives you f(x). So in this case, the question becomes: What must you differentiate in order to give you 1/x^2?

 

[Doc Al]

I have done derivatives before, but never antiderivatives before and am not sure. Just giving a wild guess would be, x ?

There's nothing wrong with guessing an antiderivative, but then you must immediately check if your guess was correct by taking the derivative. Try it.