Finding Electric Field at r>= R with Four Point Charges

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SUMMARY

The electric field at a distance r ≥ R from four point charges, consisting of two positive charges (+q) and two negative charges (-q), is zero according to Gauss's law, provided the charges are enclosed within a surface. The net charge inside the Gaussian surface is zero, leading to zero electric flux. However, in cases of non-symmetric charge distributions, such as an electric quadrupole, the electric field can be non-zero outside the enclosing surface. This highlights the importance of charge configuration when applying Gauss's law.

PREREQUISITES
  • Understanding of Gauss's law and its mathematical formulation
  • Familiarity with electric field concepts and charge distributions
  • Knowledge of electric flux and its relation to charge
  • Basic principles of electrostatics, particularly regarding point charges
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  • Study the implications of Gauss's law in non-symmetric charge distributions
  • Learn about electric quadrupoles and their electric field characteristics
  • Explore advanced applications of Gauss's law in different geometries
  • Investigate the relationship between electric field and electric potential
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Students and professionals in physics, particularly those studying electrostatics, electrical engineers, and anyone interested in the behavior of electric fields in complex charge systems.

bigplanet401
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Homework Statement



Four point charges, two of charge +q and two of charge -q, are placed in an arbitrary way. Each charge is located a distance r_i < R from the origin. Find the electric field at r >= R.

Homework Equations



According to Gauss's law, the field should be zero, right?

[tex] \oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_\text{inside}}{\epsilon_0} = 0 \, .[/tex]


The Attempt at a Solution



I'm thinking the electric field should be zero outside any surface that totally encloses the charges. Is this the correct reasoning?
 
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bigplanet401 said:

Homework Statement



Four point charges, two of charge +q and two of charge -q, are placed in an arbitrary way. Each charge is located a distance r_i < R from the origin. Find the electric field at r >= R.

Homework Equations



According to Gauss's law, the field should be zero, right?

[tex] \oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_\text{inside}}{\epsilon_0} = 0 \, .[/tex]

The Attempt at a Solution



I'm thinking the electric field should be zero outside any surface that totally encloses the charges. Is this the correct reasoning?

My reasoning is that the FLUX should be zero because the field lines corresponding to 2q+ are pointing out and the 2q- field lines point in. Mathematically, this just means [tex]Q_{net in}=0[/tex] This means that the right side of Gauss's law zeros out so the flux is zero.
 
Last edited:
Where did you get this question? You cannot use Gauss' law in general to calculate the electric field unless you have some rather strong symmetries (usually planar, cylindrical, or spherical).

The electric quadrupole is an example of using the four charges in your question while producing a nonzero electric field at r>=R. An electric quadrupole is formed by alternating positive and negative charges arranged on the corners of a square. That square could certainly lie within a sphere of radius R. But the electric field of an electric quadrupole is certainly not zero outside that sphere. Check out the electric quadrupole section of wikipedia's page on quadrupole.

As AdkinsJr says, if you were asked for the electric FLUX over any closed surface lying in the region r>=R then the answer would definitely be zero.
 

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