How Do You Calculate Electric Field and Potential at Specific Points?

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In summary, the conversation discussed finding the magnitude of the electric field at the 4th corner of a square with three positive charges, using the Couloumb Constant and electric field strength equation. The Pythagorean Theorem was also mentioned. In addition, the conversation also covered finding the electric potential at the 4th point due to charges placed on the corners of a rectangle, using the Couloumb Constant and force equations. The concept of adding vectors to get the total force per unit charge was also mentioned.
  • #1
Tsar_183
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I have two questions, if i may ask. Really confusing stuff.

1. Three positive charges are arranged as shown (diagram is of a square). The Couloumb Constant is 8.99x10^9 N*m^2/C^2.

View attachment Nhein.bmp

Find the magnitude of the electric field at the 4th corner. Answer in units of N/C.


I'm pretty sure i need these equations for this problem.

Pythagorean Theorem:
z^2 = x^2 + y^2 --> or for a square: z^2 = (2d^2)

Electric Field Strength Equation:
E = (K)(q)/(d)^2
E = Electric Field Strength
K = Couloumbs Constant
q = Charge
d = Distance

3 Directions:
X - Direction
Y - Direction
(X,Y) or Z - Direction

So I did this with my vectors -
E(x) = (8.99 x 10^9 N*m^2/C^2)(1.0 nC x 10^-9 C/nC)/(.10m)^2
E(y) = (8.99 x 10^9 N*m^2/C^2)(3.0 nC x 10^-9 C/nC)/(.10m)^2
E(z) = (8.99 x 10^9 N*m^2/C^2)(1.0 nC x 10^-9 C/nC)/(SQRT((.10m)^2 +
(.10m)^2))

However i am unsure on the 3rd Vector, and on what to do next, do i add the 3 parts together to find the total or what?


2. Consider charges placed on the corner of a rectangle: let K = 8.98755 N*m^2/C^2 and g = 9.8 m/s^2.

View attachment Yen-Xi.bmp

Find the Electric Potential at the 4th point due to the grouping of charges at the other corners of the rectangle. Answer in units of V.

OK, well yeah here is my equations that i thought might help:

F = (m)(g)
m = mass
g = gravity due to acceleration

V = (E)(d) = (F)(d)/(q)
V = Volts
E = Electric Field Strength
F = Force
d = distance
q = charge

E = (K)(q)/(d)^2
E = Electric Field Strength
K = Couloumbs Constant
q = Charge
d = Distance

3 Directions:
X - Direction
Y - Direction
(X,Y) or Z - Direction

Well i thought about using the first equation to find the forces of each vector, but that didnt work out as i have no mass to use. Was the gravity of acceleration just a trick piece to hinder me? Haha I am unsure on that. If i use the second equation as is, I don't have a force, and the third equation will give me Electric Field Strength. Is there a mass that i need to answer this question? Force is my biggest issue here.
 
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  • #2
to answer part of your question...

The field E, is just the force per unit charge. You add vectors of force to get the total force, so you add all the E vectors together to get the total force per unit charge at r.

The total force on a charge (if any) at r, is charge*force per unit charge, which is summarized as F=qE

Also- it's probably best not to use z as notation for your vector in the diagonal direction. x,y,z almost always implies that z is orthogonal to x and y.
 
  • #3
that should help, thanks :)
 

Related to How Do You Calculate Electric Field and Potential at Specific Points?

1. What is Coulomb's law?

Coulomb's law is a fundamental law of electrostatics that describes the relationship between two charged particles. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. How is Coulomb's law related to electric fields?

Coulomb's law is related to electric fields because it describes the force between two charges in terms of the electric field created by those charges. The electric field is a vector field that represents the force per unit charge at any given point in space.

3. What is the unit of electric charge in Coulomb's law?

The unit of electric charge in Coulomb's law is the Coulomb (C). One Coulomb is defined as the amount of charge that passes through a point in one second when there is a current of one ampere.

4. How is Coulomb's law used to solve problems?

Coulomb's law can be used to solve problems involving the force between two charged particles. To do so, you must first identify the charges and their distances from each other. Then, you can use the equation F = k*q1*q2/r^2, where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

5. What is the difference between Coulomb and Volt?

Coulomb and Volt are two different units of measurement in the SI system. Coulomb is a unit of electric charge, while Volt is a unit of electric potential or voltage. They are related by the equation V = W/q, where V is voltage, W is work, and q is charge. In other words, one Volt is equal to one Joule per Coulomb.

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