# Dependence of an electric field on distance

## Homework Statement

A proton located several proton diameters away from a small charged object carrying charge q is subject to an electric field of magnitude E. As the proton moves a distance d along the x axis away from the object, the electric field magnitude drops to E/4.

If the charged object had instead been an electric dipole (with a charge of magnitude q on each end) oriented in the z direction, by what factor would the initial electric field magnitude E have changed as the proton receded by distance d along the x axis?

The electric field magnitude will be increased to 2 times the original field magnitude .
The electric field magnitude will be decreased to 1/8 of the original field magnitude .
The electric field magnitude will be increased to 8 times the original field magnitude.
The electric field magnitude will be decreased to 1/2 of the original field magnitude.
The electric field magnitude will be decreased to 1/4 of the original field magnitude .
The electric field magnitude will be increased to 4 times the original field magnitude .

E = kq / r^2

## The Attempt at a Solution

E / 4 = 2 kq / d^2

So, after setting up this equation I thought that the electric field magnitude will be increased to 2 times the original magnitude. This answer was incorrect. What am I doing wrong..?

Take a look at your equation for the electric field. Does it depend on the probe charge that you put on the space?

Take a look at your equation for the electric field. Does it depend on the probe charge that you put on the space?

it depends on other variables as well, right?

it depends on other variables as well, right?
It depends on the distance from the source charge and in the charge of the source. So, does the change in the electric field depends on the probe charge?

It depends on the distance from the source charge and in the charge of the source. So, does the change in the electric field depends on the probe charge?

it depends on the distance from the source as well

vela
Staff Emeritus
Homework Helper

E = kq / r^2

## The Attempt at a Solution

E / 4 = 2 kq / d^2
How did you come up with this equation? You should think about what the relevant equation you listed is for. Does it apply here?

DaveE
Gold Member
Remember that an electric dipole is a positive charge on one side and a negative charge on the other side.

vela
Staff Emeritus
Homework Helper
Take a look at your equation for the electric field. Does it depend on the probe charge that you put on the space?
You've misunderstood the problem.

You've misunderstood the problem.
You are right. I'm very sorry if I have caused any confusion.

How did you come up with this equation? You should think about what the relevant equation you listed is for. Does it apply here?
the equation I listed is for electric fields. My thought was since their are 2 charges I should add a 2 in front of the q. I guess their is some new equation I wasn't taught...?

vela
Staff Emeritus
Homework Helper
That equation is for the electric field of a point charge, so it doesn't apply for a dipole, which is what you have here.

• downbra
That equation is for the electric field of a point charge, so it doesn't apply for a dipole, which is what you have here.
after some digging, is this the right equation? E=kqa / z^3

vela
Staff Emeritus