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

In summary: That's because the electric field is at all points perpendicular to the surface. But you are not being asked for the electric FLUX, you are being asked for the electric FIELD.In summary, according to Gauss's law, the electric field outside any surface that totally encloses the charges should be zero. However, this is not always the case, as seen with the example of an electric quadrupole where the electric field at r>=R is nonzero. This is because Gauss's law can only be used to calculate the electric field in cases with strong symmetries, such as planar, cylindrical, or spherical. Therefore, the electric field for the four point charges given in the problem cannot be determined without further information.
  • #1
104
0

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?
 
Physics news on Phys.org
  • #2
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:
  • #3
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.
 

1. How do I find the electric field at a point outside of four point charges?

To find the electric field at a point outside of four point charges, you can use the electric field equation: E = kq/r^2, where k is the Coulomb's constant, q is the charge of each point charge, and r is the distance between the point charge and the point where you are calculating the electric field. You will need to calculate the electric field for each point charge and then add them together vectorially to find the total electric field at the point outside of the four point charges.

2. Can I use the superposition principle to find the electric field at a point outside of four point charges?

Yes, you can use the superposition principle to find the electric field at a point outside of four point charges. This principle states that the total electric field at a point is equal to the vector sum of the electric fields created by each individual point charge. So, you can calculate the electric field for each point charge and then add them together vectorially to find the total electric field at the point outside of the four point charges.

3. How do I determine the direction of the electric field at a point outside of four point charges?

To determine the direction of the electric field at a point outside of four point charges, you can use the principle of superposition and vector addition. The direction of the electric field at a point is the same as the direction of the vector sum of the electric fields created by each individual point charge. You can use trigonometry and the known directions of each individual electric field to find the direction of the total electric field at the point.

4. Is the electric field at a point outside of four point charges always positive?

No, the electric field at a point outside of four point charges can be either positive or negative depending on the magnitude and direction of the individual electric fields created by each point charge. If the individual electric fields are all pointing in the same direction, the total electric field will be positive. However, if the individual electric fields are pointing in different directions, the total electric field can be negative.

5. Can I use the electric field equation to find the electric field at a point inside of four point charges?

No, the electric field equation E = kq/r^2 can only be used to find the electric field at a point outside of four point charges. Inside the four point charges, the electric field is not constant and cannot be calculated using this equation. To find the electric field inside the four point charges, you will need to use Gauss's Law or the electric field due to a continuous charge distribution formula.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
229
  • Introductory Physics Homework Help
Replies
5
Views
596
  • Introductory Physics Homework Help
Replies
4
Views
921
  • Introductory Physics Homework Help
Replies
2
Views
933
  • Introductory Physics Homework Help
Replies
3
Views
142
  • Introductory Physics Homework Help
Replies
2
Views
923
  • Introductory Physics Homework Help
Replies
3
Views
700
  • Introductory Physics Homework Help
Replies
26
Views
380
  • Introductory Physics Homework Help
Replies
11
Views
249
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top