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

  • Thread starter Thread starter Tsar_183
  • Start date Start date
  • Tags Tags
    Volt
Click For Summary
SUMMARY

This discussion focuses on calculating the electric field and potential at specific points due to multiple charges. The Coulomb Constant is specified as 8.99 x 109 N*m2/C2. Key equations include the Electric Field Strength Equation, E = (K)(q)/(d)2, and the Electric Potential Equation, V = (E)(d) = (F)(d)/(q). The participants clarify the need to sum vector components to find the total electric field and address confusion regarding the use of mass in potential calculations.

PREREQUISITES
  • Understanding of Coulomb's Law and the Coulomb Constant
  • Familiarity with vector addition in physics
  • Knowledge of electric field and electric potential concepts
  • Basic proficiency in using the Pythagorean Theorem for distance calculations
NEXT STEPS
  • Study vector addition in electric fields to accurately calculate resultant forces
  • Learn about the implications of charge distribution on electric potential
  • Explore the relationship between electric field strength and force per unit charge
  • Investigate the role of mass in electric potential calculations and its relevance
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone involved in electrostatics or electric field calculations will benefit from this discussion.

Tsar_183
Messages
6
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
that should help, thanks :)
 

Similar threads

Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
989
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 5 ·
Replies
5
Views
19K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K