Electric Dipole: Calculate Electric Field at a Distance

[vipertongn]

Homework Statement
Two point charges likes those in the figure below are called an electric dipole. Show that the electric field at a distant point along the x-axis is given by $$E_{x}=\frac{4k_{e}qa}{x^3}$$
Figure: http://img300.imageshack.us/my.php?image=58ag9.png

Homework Equations

Electric field equation: $$E=\frac{k_{e}q}{r^2}$$

The Attempt at a Solution

I know that the total electric field at some point equals the vector sum of the electric fields of both charges. So...

-kq/r^2+kq/r^2?

From the solutions it puts in x-a and x+a for r values (x+a was orignally x-(-a)). I want to know why its subtracting the vector.

 

[berkeman]

Homework Statement
Two point charges likes those in the figure below are called an electric dipole. Show that the electric field at a distant point along the x-axis is given by $$E_{x}=\frac{4k_{e}qa}{x^3}$$
Figure: http://img300.imageshack.us/my.php?image=58ag9.png

Homework Equations

Electric field equation: $$E=\frac{k_{e}q}{r^2}$$

The Attempt at a Solution

I know that the total electric field at some point equals the vector sum of the electric fields of both charges. So...

-kq/r^2+kq/r^2?

From the solutions it puts in x-a and x+a for r values (x+a was orignally x-(-a)). I want to know why its subtracting the vector.

Did you check the signs on the q's?

 

[nlightner]

I would approach this problem by first recognizing that an electric dipole consists of two equal but opposite charges separated by a distance (2a in this case). This means that the net charge of the dipole is zero, and thus the electric field at a distant point along the x-axis should also be zero.

However, when we calculate the electric field at a point due to a single charge, we use the equation E = kq/r^2, where r is the distance from the charge to the point. In this case, we have two charges, so the distance from each charge to the point will be different (x+a for the positive charge and x-a for the negative charge).

To find the total electric field at the distant point along the x-axis, we need to consider the vector sum of the individual electric fields. Since the electric field is a vector quantity, we need to take into account both magnitude and direction. Since the positive and negative charges are separated by a distance of 2a, their electric fields will have equal magnitude but opposite direction. This means that the two electric fields will cancel each other out along the x-axis, resulting in a net electric field of zero.

So why does the equation given in the problem have a non-zero value for the electric field at a distant point along the x-axis? This is because the equation is not giving the total electric field at that point, but rather the electric field due to one of the charges. In this case, the equation is giving the electric field due to the positive charge, which is located at a distance of x+a from the point. This means that the equation is only giving us the magnitude of the electric field, and we need to take into account the direction as well.

To determine the direction of the electric field, we can use the right-hand rule. Point your thumb in the direction of the positive charge (towards the right), and your fingers will curl in the direction of the electric field. Since we are interested in the electric field at a point along the x-axis, we can see that the electric field will be directed towards the positive charge. This means that the electric field at a distant point along the x-axis will be in the positive x-direction.

In summary, the equation given in the problem is not giving the total electric field at a distant point along the x-axis, but rather the electric field due to one of the charges. To find the total electric field