# Flux due to dipole

## Homework Statement

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)

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## Homework Statement

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

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

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

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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.