Calculating the electric field

In summary, the question asks for the strength and direction of the electric field at a specific position indicated by a dot in a figure. The equations provided in the textbook only apply if the two charges in the dipole are much closer together than the distance from the dipole. Using Coulomb's law and superimposing the two electric fields can provide the correct answer. The incorrect calculations using the equations from the textbook and Coulomb's law are due to incorrect values being used.
  • #1
aliaze1
174
1

Homework Statement



What are the strength and direction of the electric field at the position indicated by the dot in the figure?

knight_Figure_26_03.jpg


Homework Equations



Edipole = ~ [1/(4πε0)] * [2p/r3 ]
on the axis of an electric dipole

Edipole = ~ [-1/(4πε0)] * [p/r3 ]
in the plane perpendicular to an electric dipole

The Attempt at a Solution



Which equation should I use??

Thanks
 
Last edited:
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  • #2
Neither. Your equations only apply if the two charges in the dipole are much closer together than the distance r from the dipole.
 
  • #3
Avodyne said:
Neither. Your equations only apply if the two charges in the dipole are much closer together than the distance r from the dipole.

what formula would i use?
 
  • #4
my textbook uses these forumulas...so i tried the problem using them

so here is my process:

p = qs

s=0.1
q=1*10-9
p=1*10-10

1/(4πε0) = 9*109

r=0.05m

plugging everything into the second equation {[-1/(4πε0)] * [p/r3 ]}, i get -7200, but this is incorrect
 
  • #5
aliaze1 said:
what formula would i use?

Why not just use coulomb's law, and superimpose the two electric fields?
 
  • #6
nicksauce said:
Why not just use coulomb's law, and superimpose the two electric fields?

good idea..this was my original approach, which didn't work for some reason

E=[1/(4??0)]*[q/r2]

using the two charges:

q1=1*109
q2=-1*109

and their respective distances:

r1=0.05
r2=0.01250.5 (square root)

and doing all calculations, and then adding the two charges (3600 and -720) gives me 2880, which is incorrect...
 
  • #7
lol so yea...umm...any help?
 

1. What is an electric field?

An electric field is a physical quantity that describes the influence that a charged particle has on other charged particles in its vicinity. It is a vector quantity, meaning it has both magnitude and direction.

2. How is the electric field calculated?

The electric field is calculated by dividing the force exerted on a charged particle by its charge. The formula for calculating electric field is E = F/q, where E is the electric field, F is the force, and q is the charge.

3. What is the unit of measurement for electric field?

The unit of measurement for electric field is Newtons per Coulomb (N/C) in the SI (International System of Units) system. In the CGS (Centimeter-Gram-Second) system, the unit is dynes per statcoulomb (dyn/cm2).

4. How does distance affect the electric field?

The electric field is inversely proportional to the square of the distance between the charged particle and the point where the field is being measured. This means that as the distance increases, the electric field decreases.

5. Can the electric field be negative?

Yes, the electric field can be negative. This indicates that the direction of the field is opposite to the direction of the force on a positive test charge. A positive electric field indicates that the force and field are in the same direction.

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