Electric Flux and Gauss' Law of point charges

In summary, the problem involves finding the total electric flux through a spherical surface centered at the origin and with radius r = 1.45 m due to two point charges, q_{1} = 4.15 \times 10^-6 and q_{2} = -5.80 \times 10^-6 C, located at x = 1.80 m and y = 1.10 m respectively. The calculation involves using the equation \phi_{E} = \frac{Q_{enclosed}}{\epsilon_{0}} and converting units to arrive at a final answer of -6.55 \times 10^{5} {\rm N \cdot m}^{2}/ \rm C. The mistake
  • #1
robbondo
90
0

Homework Statement



A point charge [tex]q_{1} = 4.15 \times 10^-6 [/tex] is located on the x-axis at x = 1.80 m, and a second point charge [tex]q_{2} = -5.80 \times 10^-6[/tex] C is on the y-axis at y = 1.10 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 = 1.45 m?
Take the permittivity of free space to be [tex] 8.85 \times 10^{-12}\:{\rm C}^{2}/{\rm N \cdot m}^{2}.[/tex]


Homework Equations



[tex] \phi_{E} = \frac{Q_{enclosed}}{\epsilon_{0}} [/tex]




The Attempt at a Solution



I just divided the one charge inside the sphere by epsilon naught.

[tex] \frac{q_{2}}{\epsilon_{0}} [/tex]

So I get, [tex] -6.55 \times 10^{5} [/tex]

I'm sure I screwed something obvious up, any suggestions?
 
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  • #2
robbondo said:

Homework Equations



[tex] \phi_{E} = \frac{Q_{enclosed}}{\epsilon_{0}} [/tex]

The Attempt at a Solution



I just divided the one charge inside the sphere by epsilon naught.

[tex] \frac{q_{2}}{\epsilon_{0}} [/tex]

So I get, [tex] -6.55 \times 10^{5} [/tex]

I'm sure I screwed something obvious up, any suggestions?

Why are you sure you messed up? Show the calculation you made and also be sure to show your units.
 
  • #3
I know I'm wrong because this hw's online and I got it wrong, and I loose points for every wrong answer suckily. Well the units for q were nanocoulombs which I changed to coulombs and then epsilon naught is [tex] {\rm C}^{2}/{\rm N \cdot {m}}^{2} [/tex]. So they cancel out do give [tex] {\rm N \cdot m}^{2}/ \rm C} [/tex]

Calculation was [tex] \frac{-5.8 \cdot 10^{-12}}{8.85 \cdot 10^{-12}} [/tex]

crap... I used the wrong changing of units it [tex] 1 \cdot 10^{-9} [/tex] coulombs per nanocoulombs... not -6.

Thanks solo.
 
  • #4
robbondo said:
crap... I used the wrong changing of units it [tex] 1 \cdot 10^{-9} [/tex] coulombs per nanocoulombs... not -6.

Thanks solo.

Well, that was easy on me... As a suggestion, when you present a problem in the forum, type it in exactly as it appeared originally. That would have made the SI prefix error easy to spot. Your method was correct!
 

1. What is electric flux?

Electric flux is a measure of the flow of electric field through a given surface. It is defined as the dot product of the electric field and the perpendicular area vector of the surface.

2. What is Gauss' Law of point charges?

Gauss' Law states that the electric flux through a closed surface is proportional to the enclosed charge. This means that the total electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.

3. How is electric flux related to Gauss' Law?

Gauss' Law can be used to calculate the electric flux through a closed surface by considering the enclosed charge and the permittivity of free space. It is based on the principle that electric field lines begin and end on charges, and the total number of field lines is proportional to the amount of charge enclosed.

4. Can electric flux be negative?

Yes, electric flux can be negative. This occurs when the electric field and the perpendicular area vector of the surface are in opposite directions, resulting in a negative dot product. This can happen, for example, when the electric field is directed into the surface.

5. How can Gauss' Law be used in practical applications?

Gauss' Law can be used to calculate the electric field at any point outside of a charged sphere or other symmetrical charge distributions. It is also used in the design and analysis of capacitors, which are used in many electronic devices such as computers and cell phones.

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