# Electric flux through a surface placed between

electric flux through a surface placed between....

## Homework Statement

a circle of radius R is kept in the middle of two opposite point charges of Q,-Q.distance between the charges is A, circle is kept at A/2 . what is the electric flux through the circle?

∫E.dA=ø

## The Attempt at a Solution

E at the circle is due to both the charges which adds up.

Last edited:

Use Gauss's Law.

PeroK
Homework Helper
Gold Member
2020 Award
Gauss's Law applies to closed surfaces.

Gauss's Law applies to closed surfaces.
Sorry, I forgot to mention that I was speaking of this form of Gauss's Law:

$ø = kQ*$(solid angle made by the surface at the point)

Sorry, I forgot to mention that I was speaking of this form of Gauss's Law:

$ø = kQ*$(solid angle made by the surface at the point)
E is not uniform throughout the circular surface.an extra electric flux enters the conical boundary of the solid angle you mentioned due to the effect of the other negative charge which leaves the solid angle region through the circle adding to the flux through the circle.so one of the methods is to evaluate the integral ∫E.dA,

PeroK
Homework Helper
Gold Member
2020 Award
E is not uniform throughout the circular surface.an extra electric flux enters the conical boundary of the solid angle you mentioned due to the effect of the other negative charge which leaves the solid angle region through the circle adding to the flux through the circle.so one of the methods is to evaluate the integral ∫E.dA,
Electric fields can simply be added. So, you can calculate the flux for the positive charge, as suggested, then it's the same again for the negative charge. Using the solid angle will give you the same result as integrating the field over the circular surface.

E is not uniform throughout the circular surface.an extra electric flux enters the conical boundary of the solid angle you mentioned due to the effect of the other negative charge which leaves the solid angle region through the circle adding to the flux through the circle.so one of the methods is to evaluate the integral ∫E.dA,
The effect of one over the other can be neglected by virtue of superposition principle. Flux due to each can be calculated and then added.