Electric Field of Equilateral Triangle of Point Charges

In summary, a point charge of q = 2.7 x 10^-6 C is placed at each corner of an equilateral triangle with sides of .11 m. The magnitude of the electric field at the midpoint of any of the 3 sides of the triangle is 16,007,407 N/C. This is because the two forces acting at this point cancel out, leaving only the contribution from the charge on the opposite vertex. It is helpful to draw a figure to visualize the problem and determine the direction of the electric fields from each charge.
  • #1
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Homework Statement


A point charge q = 2.7 x 10^-6 C is placed at each corner of an equilateral triangle with sides of .11 m in length. What is the magnitude of the electric field at the midpoint of any of the 3 sides of the triangle.


Homework Equations


F = kqq/r^2
E = F/q



The Attempt at a Solution


F = k(2.7x10^-6)^2/.055^2
F = 21.66 N

Since there are two forces acting at this point
2 x 21.66 = 43.22 N is the total force.

E = F/q
43.22 N/2.7x10^-6C
= 16007407 N/C
 
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  • #2
the intensity due to the 2 charges on the same side of the triangle ( where d point is taken) cancel out since the charges are of same nature ( forces act in opposite directions). so the net field is contributed only by the charge on opposite vertex.
 
  • #3
I always believe that when you have a problem such this, it is better to draw a simple figure to be sure that what you are doing is right ..

so for your problem see the following figure:

http://img28.imageshack.us/img28/1414/64786042.jpg


figure (1) represents your problem, you have identical charges at the croners of the triangle..

lets consider the situation (figure (2) ) where we want to find the field at the midside (the one on the left) .. notice carefully the directions of the fields from each charge , what can you conclude?
 
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1. What is an electric field?

An electric field is a physical quantity that describes the influence that charged particles have on each other. It is a vector field, meaning it has both magnitude and direction, and is created by electric charges.

2. How is electric field strength measured?

Electric field strength is measured in units of volts per meter (V/m). It can be measured using a device called an electric field meter, which measures the strength of the electric field at a particular point in space.

3. What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude and distance of the charges creating the field. The strength also depends on the medium in which the charges are located, as different materials have different abilities to conduct electric fields.

4. How does an electric field interact with particles?

An electric field interacts with particles by exerting a force on them. If a particle has a charge, it will experience a force in the direction of the electric field if it is positive, and in the opposite direction if it is negative. The strength of the force depends on the charge of the particle and the strength of the electric field.

5. What are some real-world applications of electric fields?

Electric fields have many practical applications, such as in electronic devices like computers and televisions. They are also used in medical equipment like MRI machines and defibrillators. Electric fields are also used in energy production, such as in hydroelectric power plants and solar panels.

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