Electric Field Due to Point Charges?

  • Thread starter ylem
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  • #1
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Hello!

Just hoping for a bit of help on a homework problem I'm doing at the moment...

Basically this is the question:

Three point charges are placed along the y axis: a charge q at y=a, a charge -2q at the origin and a charge q at q=-a. Such an arrangement is called an electric quadrupole.
a. Find the magnitude and direction of the electric quadrupole at points along the positive x-axis.
b. Use the binomial expansion to find an approximate expression for the electric field valid for x>>a. Contrast this behaviour to that of the electric field of a dipole.


I've tried to do this loads of different ways...

I've basically used E = 1/4pi...q/r^2 (didn't know how do get a symbol for the permittivity of free space!) and then for the vector have put (xi - aj) for q at a and so on... But then when I add them up to come up with a total E I just get something that looks really messy - although it is to the power of a half and could be expanded using the binomial, it just looks wrong...

Any input on how I could try to solve the question would be greatly appreciated!

Thanks, Sam(antha)
 

Answers and Replies

  • #2
siddharth
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ylem said:
Hello!
Just hoping for a bit of help on a homework problem I'm doing at the moment...
Basically this is the question:
Three point charges are placed along the y axis: a charge q at y=a, a charge -2q at the origin and a charge q at q=-a. Such an arrangement is called an electric quadrupole.
a. Find the magnitude and direction of the electric quadrupole at points along the positive x-axis.
b. Use the binomial expansion to find an approximate expression for the electric field valid for x>>a. Contrast this behaviour to that of the electric field of a dipole.

I've tried to do this loads of different ways...
I've basically used E = 1/4pi...q/r^2 (didn't know how do get a symbol for the permittivity of free space!) and then for the vector have put (xi - aj) for q at a and so on... But then when I add them up to come up with a total E I just get something that looks really messy - although it is to the power of a half and could be expanded using the binomial, it just looks wrong...
Any input on how I could try to solve the question would be greatly appreciated!
Thanks, Sam(antha)
What you did seems to be right. Just find the vector sum of the Electric field due to the three charges. The y-component of the field due to the charge at y=a and y=-a should cancel. So along the x-axis, your net Field should also point along the x direction.
If you post and show exactly where you are having difficulties (ie, which equations) it will be easier to help.
 
Last edited:
  • #3
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Thanks!

This is where I'm getting confused though, because if the y components cancel then when I find the magnitude of the field I will just be squaring the x components, so I won't have anything to the power of a half to expand using the Binomial? :confused:
 
  • #4
siddharth
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You will have a power of 3/2. When you take the x-component due to the 2 charges at y=a and y=-a, the field will be
[tex] \frac{q \cos\theta }{4 \pi \epsilon (a^2 + x^2)} [/tex]
So find [tex] \cos \theta [/tex] and substiute to find the Field.
So you will be able to use your approximation when x>>a
 
Last edited:

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