Derive an expression for the flux through a sphere

[phosgene]

Homework Statement

Three charges with values + 20 μC, -10 μC and -10 μC are placed at the corners of a 0.10m square, as shown.

c) Consider a sphere of radius 0.3m which totally encloses all three charges. Using Gauss's law derive an expression for the flux through this surface.

Homework Equations

Electric flux = ƩQ/ε₀

The Attempt at a Solution

I don't get it, isn't Gauss's law by itself an expression for finding the electric flux (which I think should be 0)? I think it may have something to do with the non-uniform distribution of charge making Gauss's law by itself inadequate for the calculation, but I really have absolutely no idea what I'm supposed to do.

 

[Mindscrape]

The beauty of Gauss's law is that it doesn't care about where the charge is, only that you have the right total charge. So yes, you got it right, the flux is 0. If your total charge was not 0, however, the problem would be a bit trickier (trickier in that you'd have to think about what Gaussian surface to use).

 

[phosgene]

The beauty of Gauss's law is that it doesn't care about where the charge is, only that you have the right total charge. So yes, you got it right, the flux is 0. If your total charge was not 0, however, the problem would be a bit trickier (trickier in that you'd have to think about what Gaussian surface to use).

Thanks for the reply, but are you sure there's nothing else I need to do? The wording of the question made me think that I had to do more than just calculate a value for the flux.

 

[Mindscrape]

$$\Phi = \oint \mathbf{E} \cdot d \mathbf{A} = Q_{enc}/\epsilon_0$$
Though I'm sure it can be done by making holes or something, we don't really want to think about that dot product, and we can just completely ignore evaluating the left hand side of this equation because the right hand side is most definitely 0. I dunno, I guess the question was ill-formed or it was trying to mess with your mind.

 

[asteriatic]

I would approach this problem by first understanding the concept of electric flux and Gauss's law. Electric flux is a measure of the flow of electric field through a given surface, and Gauss's law relates this flux to the enclosed charge by stating that the flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

In this problem, we have three point charges located at the corners of a square and we are interested in finding the flux through a spherical surface that encloses all three charges. To do this, we can use Gauss's law by first choosing a Gaussian surface that is a sphere with a radius of 0.3m. This surface will enclose all three charges and will be symmetric with respect to the charges.

Next, we need to calculate the electric field at any point on this Gaussian surface due to the three charges. This can be done by using the principle of superposition, which states that the total electric field at any point is the vector sum of the individual electric fields due to each charge. In this case, the electric field at any point on the Gaussian surface will be the sum of the electric fields due to the three point charges.

Once we have the electric field at any point on the Gaussian surface, we can calculate the flux through the surface by taking the dot product of the electric field and the area vector of the surface. This will give us the flux through the entire surface, which we can then use in Gauss's law to find the total enclosed charge.

Therefore, the expression for the flux through the spherical surface would be:

Electric flux = (Electric field due to +20 μC charge) x (area of spherical surface) + (Electric field due to -10 μC charge) x (area of spherical surface) + (Electric field due to -10 μC charge) x (area of spherical surface)

Once we have this expression, we can use Gauss's law to find the total enclosed charge and thus calculate the flux through the spherical surface. This approach takes into account the non-uniform distribution of charge and allows us to find the electric flux through the surface correctly.