Calculating Electric Field Strength at a Corner with No Charge

In summary, the conversation discusses a problem with a practice test and previous homework problem in an electrostatics class. The problem involves finding the electric field strength at a corner with no charge in a square with charges at three corners. The conversation mentions using the equation E=(kq)/(r^2) to break the problem into vectors and determining the direction using symmetry. The final answer should be written in unit vector form, such as (x-component)i+(y-component)j.
  • #1
600burger
64
0
I'm encountering a problem with a practice test and previous homework problem in my electrostatics class.

Charges exist at three corners of a square as shown. Edge length = L. What is the Electric Field Strength (E) at the corner with no charge.


-3q
O------------O +2q
|
|
|
|
O-------------
+2q


so i break it into vectors using E= (k q)/(r^2) (Form of gauss) But i don't know what the final answer should look like...
 
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  • #2
Remember, the electric field is a vector - it has magnitude and direction!
 
  • #3
Using symmetry of this problem,
you should be able to determine the direction,
and [almost by inspection] write down the algebraic answer.

Hint: if you can determine the direction (using symmetry), it is advantageous to use a coordinate system that has an axis along this direction [instead of the standard x- and y- axes]. The only components you'll need to consider are those along this axis.
 
  • #4
Right a vector. So i would come out with an answer in unit vector form...like...



(x-component)i+(y-component)j. Or is there a simpiler way to write it?
 

FAQ: Calculating Electric Field Strength at a Corner with No Charge

What is Gauss's law and how does it relate to vectors?

Gauss's law is a fundamental principle in physics that relates the flux of an electric field through a closed surface to the enclosed charge. It is often written in vector form, using vector calculus to describe the electric field and surface. In essence, it states that the total electric flux through a closed surface is equal to the total enclosed charge divided by a constant known as the permittivity of free space.

How is Gauss's law used in electrostatics?

Gauss's law is used extensively in electrostatics to calculate the electric field in various situations. It is particularly useful in situations with high degrees of symmetry, as it allows for simplified calculations and can yield elegant solutions. It is also used to derive other important laws and principles in electrostatics, such as Coulomb's law and the principle of superposition.

Can Gauss's law be applied to magnetic fields?

No, Gauss's law only applies to electric fields. However, there is a similar principle for magnetic fields known as Gauss's law for magnetism, which states that the total magnetic flux through a closed surface is always zero.

What is the significance of using vectors in Gauss's law?

Vectors are used in Gauss's law because they allow for a more complete and accurate representation of the electric field. Since electric fields have both magnitude and direction, vector calculus is necessary to fully describe and calculate the flux through a surface. Additionally, using vectors allows for the application of mathematical principles such as the divergence theorem, which is crucial in solving problems involving Gauss's law.

Can Gauss's law be applied to non-uniform electric fields?

Yes, Gauss's law can be applied to non-uniform electric fields. However, in these cases, the electric field and surface must be divided into small sections, and the flux through each section must be calculated separately. The total flux can then be found by summing up the flux through each section. This method is known as the differential form of Gauss's law.

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