# Gauss's law

## Homework Statement

Four point charges, two of charge +q and two of charge -q, are placed in an arbitrary way. Each charge is located a distance r_i < R from the origin. Find the electric field at r >= R.

## Homework Equations

According to Gauss's law, the field should be zero, right?

$$\oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_\text{inside}}{\epsilon_0} = 0 \, .$$

## The Attempt at a Solution

I'm thinking the electric field should be zero outside any surface that totally encloses the charges. Is this the correct reasoning?

## Homework Statement

Four point charges, two of charge +q and two of charge -q, are placed in an arbitrary way. Each charge is located a distance r_i < R from the origin. Find the electric field at r >= R.

## Homework Equations

According to Gauss's law, the field should be zero, right?

$$\oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_\text{inside}}{\epsilon_0} = 0 \, .$$

## The Attempt at a Solution

I'm thinking the electric field should be zero outside any surface that totally encloses the charges. Is this the correct reasoning?

My reasoning is that the FLUX should be zero because the field lines corresponding to 2q+ are pointing out and the 2q- field lines point in. Mathematically, this just means $$Q_{net in}=0$$ This means that the right side of Gauss's law zeros out so the flux is zero.

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Where did you get this question? You cannot use Gauss' law in general to calculate the electric field unless you have some rather strong symmetries (usually planar, cylindrical, or spherical).

The electric quadrupole is an example of using the four charges in your question while producing a nonzero electric field at r>=R. An electric quadrupole is formed by alternating positive and negative charges arranged on the corners of a square. That square could certainly lie within a sphere of radius R. But the electric field of an electric quadrupole is certainly not zero outside that sphere. Check out the electric quadrupole section of wikipedia's page on quadrupole.

As AdkinsJr says, if you were asked for the electric FLUX over any closed surface lying in the region r>=R then the answer would definitely be zero.