How Do You Calculate the Electric Field of a Dipole at a Specific Point?

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SUMMARY

The electric field of a dipole at a specific point can be calculated using the formula E = 2qs / (4πE * r^3), where q is the charge, s is the separation distance, and r is the distance from the dipole. In this discussion, a dipole with charges of +4 nC and -4 nC, separated by 2 mm, was analyzed at a point 5 cm away. The correct calculation yields an electric field magnitude of 2.88 N/C, but the user initially neglected to cube the distance, leading to confusion. Additionally, it was confirmed that the electric field perpendicular to the dipole is indeed half of that on-axis, resulting in a value of 1.44 N/C at that position.

PREREQUISITES
  • Understanding of electric dipoles and their properties
  • Familiarity with the formula for electric fields
  • Basic unit conversion skills
  • Knowledge of the permittivity of vacuum (ε₀)
NEXT STEPS
  • Study the derivation of the electric field formula for dipoles
  • Learn about the concept of electric field lines and their representation
  • Explore the effects of varying distance on electric field strength
  • Investigate the behavior of electric fields in different geometrical configurations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields generated by dipoles.

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Homework Statement


A dipole consists of two charges +q and -q, held apart by a rod of length s, as shown in the diagram. If q = 4 nC and s = 2 mm, what is the magnitude of the electric field due to the dipole at location A, a distance d = 5 cm from the dipole?

The diagram shows point A being on axis with the dipole.

Homework Equations



The formula for on axis is 2qs/4piE*r^3, where E is the permittivity of vacuum

The Attempt at a Solution



So to solve this first I converted all units

q = 4nC = 4e-9C
s = 2mm = .002m
d = 5cm = .05m

Then I plug in the values into the formula

2*4e-9C*.002m/4pi*8.85e-12*.05m = 2.88N/C
But that seems to be the wrong answer can someone please tell me what I'm doing wrong
and also is it true that the field perpendicular to the dipole is just half of what it would be if it was on axis. If that's true than the field 5cm perpendicular to the dipole should be just half of what I come up with, in this case 1.44N/C.

Thank You for your help
 
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Oh I think I figured out where I went wrong I forgot to cube d,
 

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