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A dipole lies on the x-axis,with the positive charge on x =+d/2 and the negative charge on x=-d/2. Find the electric flux through the yz plane midway between the charges.

 A dipole lies on the x-axis,with the positive charge on x =+d/2 and the negative charge on x=-d/2. Find the electric flux through the yz plane midway between the charges.

Grade:12

1 Answers

Arun
25763 Points
2 years ago
The total flux through the y-z plane perpendicular to the axis of the electric dipole is -Q/(4eo) where Q/2 is the magnitude of each charge on the dipole and eo (usually written as epsilon sub o) is a constant describing the electric properties (permittivity) of the medium surrounding the dipole. If the medium is a vacuum, then eo = 8.85 x 10^-12. We won't worry about the units for now.

Without going into the detailed mathematics there is a simple way to determine the total flux passing through the plane passing between the two charges on the dipole. We use Gauss’s law which states that the total flux passing through a closed surface surrounding any charge is equal to the total charge inside the surface divided by eo. To treat a dipole we have to do it in two steps. first a surface surrounding the positive charge and then a surface surrounding the negative charge of the dipole.

The choice of surface is the important part of solving this problem. For the positive charge the first part of the surface is the our y-z plane. To close the surface we chose a hemisphere of very large radius (actually infinite radius). There is no flux through the hemisphere since the field of the enclosed charge decreases as r^-2 so as r increases the electric field approaches zero. Then the only flux is through the y-z plane and must equal by Gauss’s law Q/2eo and directed out of the closed surface in the -x direction. Using the same argument for the negative charge of the dipole we now use a hemisphere below the y-z plane and enclosing the negative charge with the y-z plane as the upper boundary of the closed surface. Now the only difference is that the flux is into the closed surface through the y-z plane because this charge is negative.

The total flux is then the sum of the two fluxes which are in the same direction normal to the y-z plaane in the minus x direction.

Total flux is a vector equal to -Q/(4eo) *axwhere axis the unit vector in the +x direction

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