Electric flux through a spherical surface

In summary, the problem involves three point charges near a spherical Gaussian surface of radius R = 6 cm. One charge of +3Q is inside the sphere, while the others are a distance R/3 outside the surface. The question asks for the electric flux through the surface, given that Q = 6 C. To solve this, Gauss's law for a sphere can be used, with the equation EA= Q(enclosed)/8.55e-12 and A for a sphere being 4Pi r^2. The electric field lines passing through the surface are independent of the surface area, so the electric field can be taken out of the integral, giving the equation \vec{E}A=\frac{1}{\
  • #1
Boilermaker54
1
0

Homework Statement



Three point charges are located near a spherical Gaussian surface of radius R = 6 cm. One charge (+3Q) is inside a sphere, and the others are a distance R/3 outside the surface.

What is the electric flux through this surface (Q = 6 C)

Homework Equations


I am aware of guass's law for a sphere. EA= Q(enclosed)/8.55e-12 A for sphere = 4Pi r^2


The Attempt at a Solution




I'm not sure what radius or Q charge should get put int ote equation, I'm still trying to understand the concepts behind these problems.
 
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  • #2
The problem tells you that the radius is 6cm and the charge inside the sphere is +3Q
 
  • #3
ill try to provide some intuition behind this type of problem.

a flux is the amount of a volume passing a point per time. in this type of problem we can use electric field lines as our volume, and the point being the surface of the sphere. if one assumes that the field lines start at the center of the sphere, no field lines stop abruptly or ones begin at any other point, then no matter the surface area of the sphere the same number or field lines are passing.
Gauss's theroem says: [tex]\int(\nabla\cdot\vec{v})\:d\tau=\int\vec{v}\cdot d\vec{a}[/tex]
since [tex]\int(\nabla\cdot\vec{E})\:d\tau=\frac{1}{\epsilon}q[/tex]
[tex]\int\vec{E}\cdot d\vec{a}=\frac{1}{\epsilon}q[/tex]

but since no matter what the change in area the same amount of electric field lines are passing through the surface area. the E is independent of the da, and can be taken out of the integral. giving u [tex]\vec{E}\int d\vec{a}=\frac{1}{\epsilon}q[/tex] or [tex]\vec{E}A=\frac{1}{\epsilon}q[/tex] A being [tex] 4\pi r^{2}[/tex]
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that was my understanding on Gauss's theorem, but if i understood it incorrectly someone correct me before i confuse this good sir.
 

1. What is electric flux through a spherical surface?

The electric flux through a spherical surface is the measure of the electric field passing through the surface. It is a measure of the number of electric field lines passing through the surface per unit area.

2. How is the electric flux through a spherical surface calculated?

The electric flux through a spherical surface can be calculated by taking the dot product of the electric field vector and the surface vector, and then multiplying it by the surface area. Mathematically, it can be represented as Φ = E · A, where Φ is the electric flux, E is the electric field, and A is the surface area.

3. What is the unit of electric flux through a spherical surface?

The unit of electric flux through a spherical surface is Nm²/C, where N represents Newtons (the unit of force), m² represents square meters (the unit of area), and C represents Coulombs (the unit of electric charge).

4. How does the electric flux through a spherical surface depend on the charge enclosed within the surface?

The electric flux through a spherical surface is directly proportional to the charge enclosed within the surface. This means that as the charge enclosed increases, the electric flux through the surface also increases.

5. Can the electric flux through a spherical surface be negative?

Yes, the electric flux through a spherical surface can be negative. This occurs when the electric field and the surface vector are in opposite directions, resulting in a negative dot product. This can happen when the enclosed charge is negative or when the electric field is directed away from the surface.

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