E field due to a finite line of charge

[Jake 7174]

Homework Statement

A finite line of uniform charge is on the line joining (4,0,0) (0,8,0) and has λ= 2μC/m find E at (0,0,0)

Homework Equations

E = Q / (4 π ε₀ r^2)

dQ = λ dl = λ dx

The Attempt at a Solution

I first notice that r will depend on 2 changing values x and y.

I write an equation for the line.
y = 8 - 2x

I use this relation to find r in terms of x
r = sqrt ( x^2 + (8 - 2x)^2 )

I set up my integral
λ / (4 π ε₀ r^2) ∫ < -x , 2x - 8> dx / [ x^2 + (8 - 2x)^2]^(3/2) from x=0 to x=4

when I evaluate I get <-2250 , -2250>

This does not seem correct to me. Based on the location of the line to the origin I would expect my x component to have a larger magnitude than the y component. I am not sure where I am going wrong.

 

[gneill]

I think perhaps you need to think about what dq is in this scenario. Since the line of charge is on an angle, from the point of view of the origin dq will not be simply λ dx.

 

[Jake 7174]

I think perhaps you need to think about what dq is in this scenario. Since the line of charge is on an angle, from the point of view of the origin dq will not be simply λ dx.

My thought was that since I expressed everything in terms of x that dx would resolve the need for dy. Apparently this is not the case. I think then dq =λ dy dx.

This still does not seem correct it would turn into ∫ ∫ dy dx which feels very wrong. I am missing some understanding. Can you nudge me?

 

[gneill]

Consider this figure:

(revised image)

Write an expression for dq in terms of dx and dy. What's dy in terms of dx for your line? (think slope).

 

[Jake 7174]

Consider this figure:
https://www.physicsforums.com/attachments/95286

Write an expression for dq in terms of dx and dy. What's dy in terms of dx for your line? (think slope).

Ok, I see dq = dy/dx.
This confuses me where to go from here. If I take the derivative of my expression with respect to x wouldn't the integral just go away? How would I evaluate it then for all dq?

 

[gneill]

No, the dq will be λ times the length of the line segment indicated. For your line, for a given dx the length of dy is ...?

 

[gneill]

I've revised the image in post #4 slightly.

 

[Jake 7174]

No, the dq will be λ times the length of the line segment indicated. For your line, for a given dx the length of dy is ...?

I don't see it. The only thing I see doesn't make much sense sqrt (dq^2 - dx^2) ?

 

[Jake 7174]

I don't see it. The only thing I see doesn't make much sense sqrt (dq^2 - dx^2) ?

don't answer yet, I didn't see your revision

 

[gneill]

What's the length of the line segment containing the charge in terms of dx and dy?

What is dy in terms of dx for your particular line? (dy = ?dx)

Substitute for dy and write the length of the line segment in terms of dx only.

 

[gneill]

don't answer yet, I didn't see your revision

Too late!

 

[Jake 7174]

What's the length of the line segment containing the charge in terms of dx and dy?

What is dy in terms of dx for your particular line? (dy = ?dx)

Substitute for dy and write the length of the line segment in terms of dx only.

Is the length of the line segment dl = sqrt(dy^2 + dx^2)?

 

[gneill]

Is the length of the line segment dl = sqrt(dy^2 + dx^2)?

Yes. Simple Pythagoras. What's dy in terms of dx for your line?

 

[Jake 7174]

Yes. Simple Pythagoras. What's dy in terms of dx for your line?

sqrt(dl^2 - dx^2) ?

 

[gneill]

No, for your line as a whole: What's dy/dx? So dy = (?)dx.

 

[Jake 7174]

No, for your line as a whole: What's dy/dx? So dy = (?)dx.

-2 (from the slope)

 

[gneill]

-2 (from the slope)

Yes, so replace dy with -2dx in the expression for dl. Then you'll have an expression for dl in terms of dx only.

So then, what's dq in terms of dx?

 

[Jake 7174]

Yes, so replace dy with -2dx in the expression for dl. Then you'll have an expression for dl in terms of dx only.

So then, what's dq in terms of dx?

sqrt(5) dx

 

[gneill]

sqrt(5) dx

Multiply that length by λ to yield the charge element.

 

[Jake 7174]

Multiply that length by λ to yield the charge element.

So from that,

E = λ sqrt(5) / (4 π ε₀ ) ∫ < -x , 2x - 8> dx / [ x^2 + (8 - 2x)^2]^(3/2) from x=0 to x=4

But if I do this I just multiply my answer by a sqrt(5). Something must be wrong with my integral.

 

[Jake 7174]

So from that,

E = λ sqrt(5) / (4 π ε₀ ) ∫ < -x , 2x - 8> dx / [ x^2 + (8 - 2x)^2]^(3/2) from x=0 to x=4

But if I do this I just multiply a sqrt(5) to my original answer. Something must be wrong with my integral.

 

[Jake 7174]

So from that,

E = λ sqrt(5) / (4 π ε₀ ) ∫ < -x , 2x - 8> dx / [ x^2 + (8 - 2x)^2]^(3/2) from x=0 to x=4

But if I do this I just multiply my answer by a sqrt(5). Something must be wrong with my integral.

Please don't leave me now. We are so close. I know there must be a problem with my integral, I just don't see it.

 

[gneill]

What makes you feel that there's something wrong with your integral?

 

[Jake 7174]

What makes you feel that there's something wrong with your integral?

Because when I evaluate it I get the magnitude of e field in x direction equal to the y. This does not seem right because E is dependent on 1/r^2. My x component should be bigger I think. My answer is

<-2250 Sqrt(5) , -2250 Sqrt(5)> N/C

 

[gneill]

I'm not seeing a problem with your integrations. I think it just turns out that the components end up being equal in magnitude.

 

[Jake 7174]

I'm not seeing a problem with your integrations. I think it just turns out that the components end up being equal in magnitude.

Ok. Ill trust he math. Thanks for your help.

 

[TSny]

My answer is

<-2250 Sqrt(5) , -2250 Sqrt(5)> N/C

I also agree with this answer.

In fact, I believe that as long as the wire extends from some point on the x-axis to some point on the y-axis (neither point at the origin), the x and y components at the origin have equal magnitudes. A somewhat surprising result!

 

[Jake 7174]

I also agree with this answer.

In fact, I believe that as long as the wire extends from some point on the x-axis to some point on the y-axis (neither point at the origin), the x and y components at the origin have equal magnitudes. A somewhat surprising result!

Very much surprising. It is for this reason I suspect it was assigned. Because I could not believe it in the first place that prompted me to make my the original post. Had I not I would have missed the sqrt(5)

 

[TSny]

For what it's worth, I get that the x and y components of the field each have a magnitude of kQ/(x_oy_o), where k is Coulomb's constant, Q is the total charge on the wire, and x_o, y₀ locate the ends of the wire on the x, y axes respectively.

 

[Jake 7174]

For what it's worth, I get that the x and y components of the field each have a magnitude of kQ/(x_oy_o), where k is Coulomb's constant, Q is the total charge on the wire, and x_o, y₀ locate the ends of the wire on the x, y axes respectively.

It adds a degree of comfort. Do you all think that my approach was reasonable. Was there a better way to approach this?

 

[TSny]

It adds a degree of comfort. Do you all think that my approach was reasonable. Was there a better way to approach this?

I think your approach was very good. The only mistake was expressing dq as λdx rather than λds.