Electric force in a dipole

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  • #1
cathb
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Hi,
I had this question in my homework:

An electric dipole consists of two equal loads and signs opposite separated by one distance which equals 2a. Show how the module of the electric force resulting on a proton (load +e) placed directly on the Y=axis at a very long distance from the dipole (y>>a) is given by Fe = 2aekq/y^3

The usual equation for the electric force is: Fe= keq/r^2

I tried to replace the distance r^2 with Pythagore: a^2 + y^2 = r^2, but i didnt find the y^3
Could anybody help me?
 

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  • #2
SammyS
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Is the dipole oriented along the x-axis or the y-axis?

Assuming the dipole is oriented along the x-axis, place charge q at x=a/2 and charge ‒q at x = ‒a/2. Find the vector sum of the force each charge in the dipole exerts on the proton. (The y-components of the two forces should cancel, leaving only the x-components.)
 
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  • #3
catsy
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Hi,

Thanks for your reply. I actually have the same homework and I'm having trouble with the same question.

Here is the image of the dipole in question. My issue is how do we actually demonstrate that the module of the electric force resulting on a proton (load +e) placed directly on the Y=axis at a very long distance from the dipole (y>>a) is given by Fe = 2aekq/y^3 (i think I'm just not getting it at all... sorry :S)




Is the dipole oriented along the x-axis or the y-axis?

Assuming the dipole is oriented along the x-axis, place charge q at x=a/2 and charge ‒q at x = ‒a/2. Find the vector sum of the force each charge in the dipole exerts on the proton. (The y-components of the two forces should cancel, leaving only the x-components.)
 

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  • #4
SammyS
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Is the dipole oriented along the x-axis or the y-axis?

Assuming the dipole is oriented along the x-axis, place charge q at x=a/2 and charge ‒q at x = ‒a/2. Find the vector sum of the force each charge in the dipole exerts on the proton. (The y-components of the two forces should cancel, leaving only the x-components.)

I didn't read closely enough.

Place charge ‒q at x=a and charge +q at x = ‒a, as in the figure.

The force, F, on the proton, due to charge ‒q is: F= ____ ? (It's a vector.)

The force, F+, on the proton, due to charge +q is: F+= ____ ? (It's also a vector.)
 
  • #5
cathb
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Thank you for your help,

in brief, i only found the vector on the x-axis (cuz as you said the y-component should be canceled) and replaced the sin by tan (because of the very long distance of the y-component of the proton, which was nearly equal to the distance between the proton and one of the chage q). So the tan would be replaced by a/y.
 
  • #6
SammyS
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Thank you for your help,

in brief, i only found the vector on the x-axis (cuz as you said the y-component should be canceled) and replaced the sin by tan (because of the very long distance of the y-component of the proton, which was nearly equal to the distance between the proton and one of the charge q). So the tan would be replaced by a/y.

If r2 = a2 + y2, then for y » a, r2 ≈ y2.

That should get you your desired result.
 
  • #7
cathb
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Thanks a lot
 
  • #8
SammyS
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You're very welcome.
 

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