Electric flux of a particle

[tag16]

Homework Statement

A particle with charge Q= 5.00 microcoulombs is located at the center of a cube of edge L= 0.100 m. In addition, six other identical charged particles -q are positioned symmetrically around Q. Determine the electric flux through one face of the cube.

Homework Equations

Flux= S E dA (the S is suppose to be an integral symbol)

The Attempt at a Solution

Flux= S E dA + S E dA

is this right so far? if so not sure what I need to do next, if not what do I need to do? thanks

 

[tiny-tim]

Hi tag16!

(have an integral: ∫ )

Flux = ∫ E dA + ∫ E dA

is this right so far? if so not sure what I need to do next, if not what do I need to do? thanks

For the charge at the centre, the flux is easy to find … just find the flux through all 6 faces, and divide by 6.

But for the other charges, yes, I think you'll have to do some integrating.

 

[tomcue]

I would say that your attempt at a solution is on the right track. However, you have not fully utilized the given information and equations to solve for the electric flux through one face of the cube.

Firstly, we need to understand what electric flux is. Electric flux is a measure of the electric field passing through a given area. Mathematically, it is represented as the dot product of the electric field (E) and the area vector (dA). In this case, we can consider the electric field due to the charged particle at the center of the cube (Q) and the six identical charged particles surrounding it (-q) as the total electric field (E) passing through the face of the cube.

Secondly, we need to calculate the total electric field (E) at the center of the cube due to the seven charged particles using the principle of superposition. This means that we need to add up the contributions of each individual charged particle to the total electric field at the center of the cube. This can be done using Coulomb's law, which states that the electric field at a point due to a point charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the two points.

Once we have the total electric field (E), we can use the given equation for electric flux (Flux= S E dA) to calculate the electric flux through one face of the cube. This involves taking the integral over the surface area of the face, which can be calculated using the given edge length (L).

To summarize, your attempt at a solution is on the right track, but you need to consider the total electric field at the center of the cube and use the given equations to calculate the electric flux through one face of the cube.