# Flux due to dipole

1. Jul 12, 2010

### venkat1989

1. The problem statement, all variables and given/known data
an electrical dipole is placed at the centre of the sphere,the electric flux due to dipole is?
i know the answer is zero but i dont find my reasons satisfactory pls post reply
(i thought if the flux are normal to the centre the cos90 is zero hence zero)

2. Jul 12, 2010

### bp_psy

Do you know Gauss 's Law? What is the charge enclosed in that sphere?

3. Jul 12, 2010

### venkat1989

hey the sphere has a charge(some 'Q') but now there is a dipole and it has certain charge........

4. Jul 13, 2010

### RoyalCat

The sphere is a mathematical construct, not a physical entity. It has no charge of its own.

Before you read on, think what the total charge of an electric dipole is, and how that affects the result you'd expect from Gauss' Law.

The $$q$$ in Gauss' Law:

$$\Phi_E=\frac{q}{\epsilon_0}$$

Is the TOTAL charge enclosed by the Gaussian surface (The charge inside the closed surface!), not the charge on the surface itself.

If you were to meticulously calculate the electric flux due to the dipole through the sphere enclosing it, using the definition of the E-field flux as: $$\Phi_E = \oint \vec E \cdot d\vec A$$ you will find the total flux to be 0.

Last edited: Jul 13, 2010
5. Jul 13, 2010

### dulrich

Gauss' law is the right way to approach this problem. One way to "understand" Gauss' law is that the "number" of field lines (i.e., flux) that cross through the boundary is proportional to the charge enclosed. For the dipole, every field line that goes out the plus charge goes back into the negative charge. So the net contribution across the whole surface is zero.