Why does the electric field around a dipole fall off as r^3?

Can someone please explain why this is true? Also, why does it even fall as as r^2 when it's not around a dipole. I just want to understand the fundamentals instead of just memorizing an equation.


3 Answers

  • 7 years ago
    Favorite Answer

    For the 1/r^2 law, it's because the flux of the electric field through a closed surface is constant. As r increases, the surface area of the sphere of radius r increases proportional to r^2 (specifically 4r^2). Therefore the flux per unit area, which is electric field, is inversely proportional to r^2.

    Now for the dipole field. For a simple example, consider a dipole of two charges and points on the line that connects those charges. At a distance, the electric fields of the two charges are opposite and nearly equal in magnitude. They are not equal, because the distance to one of the charges is slightly further. Say the distance to one of the charges is r, and the distance to the other is r + dr. Then the total electric field is proportional to 1/(r + dr)^2 - 1/r^2. The limit of that as dr -> 0 is the definition of the derivative of 1/r^2, which is -2/r^3. So the electric field from the dipole is proportional to 1/r^3.

  • Anonymous
    7 years ago

    Do it mathematically. For r>>a its proportional to 1/r^3

    where r is the distance from the mid-point of the dipole to the test charge and a is the distance between the two charges in the dipole.

    Maybe the dipoles are attracting each other so the electric field outside them is weaker than if we take single charge rather.

    The force between two dipole is proportional to 1/a^2 and and the resultant vector from the middle of the dipole to the point, we divide it three time i.e r^3.

  • Anonymous
    4 years ago

    an electric dipole saved in a uniform electric container stories a stress : fake an electric dipole saved in a uniform electric container stories a torque : real Dipole in an exterior container evaluate an eternal dipole of dipole 2nd 'p' in a uniform exterior container E, which suggests that p would not rely E nor it is brought on using E. The stress on q is qE and on -q is -qE and internet stress on dipole is 0. besides the incontrovertible fact that, the charges are separated so the stress act on distinctive factors, giving upward push to a torque (twisting stress) on the dipole. whilst internet stress is 0, torque is autonomous of the beginning. Its magnitude equals the magnitude of each stress prolonged by perpendicular distance between the two antiparallel forces. it somewhat is, magnitude of torque (t) = qE x 2a sinq In vector notationare assumed to be interior the airplane of the paper. From the above, it is got here across that an electric dipole placed in uniform electric container stories a torque which align the dipole parallel to the path of the electrical powered container.

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