Ambiguous Electric-Field magnitude

  • Thread starter Thread starter PHYHelp519
  • Start date Start date
  • Tags Tags
    Magnitude
Click For Summary
SUMMARY

The discussion centers on calculating the electric field magnitude at point P due to two charges, +Q at x = -a and -Q at x = +a, with point P located at x = +b (where b > a). The correct formula for the electric field is derived as Ep = 4kQab/(b² - a²)². However, the contributor's confusion arises from the notation used for a² and b², leading to questions about the contributions from both charges. The direction of the electric field at point P is determined to be towards the negative charge.

PREREQUISITES
  • Understanding of electric field concepts and equations, specifically E = kqr/r³.
  • Familiarity with charge interactions and their effects on electric fields.
  • Basic algebra skills for manipulating equations involving variables such as a and b.
  • Knowledge of vector direction in electric fields, particularly in relation to positive and negative charges.
NEXT STEPS
  • Review the derivation of electric field equations for point charges.
  • Study the principle of superposition in electric fields to understand contributions from multiple charges.
  • Learn about the graphical representation of electric fields and their directions.
  • Explore advanced topics in electrostatics, such as electric field lines and their implications.
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric fields and charge interactions.

PHYHelp519
Messages
1
Reaction score
0

Homework Statement


On the x axis, charge +Q lies at x = -a and charge -Q lies at x = +a. Point P lies on the axis at
x = +b, where b > a. a) Work out the electric field magnitude of P b) What is the direction of this …field?

Homework Equations


E=kqr/r3

The Attempt at a Solution


E=2kQa/(b2-a2) + 2kQa/(b2-a2)=
Ep=4kQab/(b^2 - a^2)^2. My teacher said that the solution was correct, however my work was not.
For b, I determined the direction to be to the right towards the negative charge.

I would appreciate any assistance.
 
Physics news on Phys.org
PHYHelp519 said:

Homework Statement


On the x axis, charge +Q lies at x = -a and charge -Q lies at x = +a. Point P lies on the axis at
x = +b, where b > a. a) Work out the electric field magnitude of P b) What is the direction of this …field?

Homework Equations


E=kqr/r3

The Attempt at a Solution


E=2kQa/(b2-a2) + 2kQa/(b2-a2)=
Ep=4kQab/(b^2 - a^2)^2. My teacher said that the solution was correct, however my work was not.
For b, I determined the direction to be to the right towards the negative charge.

I would appreciate any assistance.

The equation in red does not make sense. What are a2, b2? Do you mean that the contribution is the same from both charges?

ehild
 

Similar threads

Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
9
Views
2K
Replies
17
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
1
Views
2K