Electric Flux due to Two Point Charges: Analysis and Calculations

In summary, the total electric flux due to the two point charges q_1 = 3.40 {\rm nC} and q_2 = -5.80 {\rm nC} through a spherical surface centered at the origin and with radius r_1 = 0.710 {\rm m} is -2.7 x10^11 Amps. The total electric flux due to the two point charges q_1 = 3.40 {\rm nC} and q_2 = -5.80 {\rm nC} through a spherical surface centered at the origin and with radius r_2 = 1.65 {\rm m} is -9.4 x10^11 Amps. The total electric flux due
  • #1
Aserap
6
0
A point charge q_1 = 3.40 {\rm nC} is located on the x-axis at x = 1.80 {\rm m}, and a second point charge q_2 = -5.80 {\rm nC} is on the y-axis at y = 1.10 {\rm m}

What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r_1 = 0.710 {\rm m}?
What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r_2 = 1.65 {\rm m}?
What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r_3 = 2.95 {\rm m}?

i am so confused at where to start! i keep getting the wrong answer
 
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  • #2
Show what you've done so far.

Hint: Use Gauss's law.
 
  • #3
There are so many ways to use gauss's law.
i used the equation [tex]\phi[/tex] = q/ E0 for each point and then added them together to get the total. the total i found for the first part was -2.7 x10^11. But it is wrong. i am not sure if i need to take into account the location of each of the points or not.
 
  • #4
Aserap said:
i am not sure if i need to take into account the location of each of the points or not.
Of course you do. Gauss's law tells you the total flux through a closed surface in terms of the total charge within that surface. You have to know if the charges are inside or outside of your spherical surface in each case.
 
  • #5
So would i use the electric field equation to find the value of each point, where r is the radius of the sphere plus the distance of each point. then i would find the flux for both and add them together? that the only possible answer i can think about doing! thanks for helping by the way!
 
  • #6
Aserap said:
So would i use the electric field equation to find the value of each point, where r is the radius of the sphere plus the distance of each point.
No. The only equation you need is Gauss's law. All you have to do, for each of the three cases in your problem, is figure out the total charge contained inside your spherical surface. Then use Gauss's law to find the total flux through that surface.

It's easier than you think. :wink:
 
  • #7
okay, so would the answer for each section of the problem be the same answer?
 
  • #8
Aserap said:
okay, so would the answer for each section of the problem be the same answer?
Nope.

Ask yourself: Are the charges inside or outside the spherical surface? That depends on how big the spherical surface is, right?
 

FAQ: Electric Flux due to Two Point Charges: Analysis and Calculations

What is electric flux?

Electric flux is a measure of the flow of electric field through a given surface. It is represented by the symbol Φ and is measured in units of volts per meter (V/m).

How is electric flux affected by two point charges?

The presence of two point charges can affect the electric flux through a given surface by changing the strength and direction of the electric field. The electric flux is directly proportional to the magnitude of the charges and inversely proportional to the distance between them.

How do you calculate electric flux due to two point charges?

The electric flux due to two point charges can be calculated using the formula Φ = Q/ε0, where Q is the total charge and ε0 is the permittivity of free space. Alternatively, it can also be calculated by integrating the electric field over the surface in question.

What factors can affect the magnitude of electric flux?

The magnitude of electric flux can be affected by several factors, such as the distance between the two point charges, the strength of the charges, and the shape and orientation of the surface through which the flux is being measured.

How can the concept of electric flux be applied in real-world situations?

The concept of electric flux can be applied in various real-world situations, such as understanding the behavior of electric fields in circuits, determining the strength of electric fields in different materials, and predicting the movement of charged particles in electromagnetic fields.

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