# Electric flux through a surface placed between

1. Feb 5, 2014

### basheer uddin

electric flux through a surface placed between....

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
∫E.dA=ø

3. The attempt at a solution
E at the circle is due to both the charges which adds up.

Last edited: Feb 5, 2014
2. Feb 5, 2014

Use Gauss's Law.

3. Feb 5, 2014

### PeroK

Gauss's Law applies to closed surfaces.

4. Feb 6, 2014

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)

5. Feb 6, 2014

### basheer uddin

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,

6. Feb 6, 2014

### PeroK

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.

7. Feb 6, 2014