Question on dipole and electric fields

[r3dxp]

Homework Statement

Two dipoles are oriented as shown in the diagram below. Each dipole consists of two charges +q and -q, held apart by a rod of length s, and the center of each dipole is a distance d from location A. If q = 3 nC, s = 1 mm, and d = 7 cm, what is the electric field at location A?

Hint: draw a diagram and show the direction of each dipole's contribution to the electric field on the diagram (you do not have to turn in the diagram).
look at picture below:
http://www.prism.gatech.edu/~bpark6/two_dipolesA.gif

Homework Equations

E= [1/(4*pi*E)] * [q*s] * [1/r^3]
where [1/(4*pi*E)] ~= 9E9
and q = charge
s = distance between dipole in meters
r = distance between the location and the center of the dipole in meters

The Attempt at a Solution

the answer has to be in a format of < x, y, z >N/C
and i used the perpendicular axis equation for the x-axis and dipole axis equation for the y-axis and got..
<78.72, 157.43, 0> N/C , and i tried all different kinds of combination of + and - for the sign, and got it wrong. I am on 4/5 submission and this is my last submission, so if any help, it will be greatly appreciated. Thanks in advance.

 

[anathema]

Hi there,

Based on your attempt at a solution, it looks like you have the right idea but may have made a small mistake in your calculations. I would suggest double checking your calculations and also taking into account the distance between the center of the dipole and location A, which is d in this case. Remember that the electric field from each dipole will have components in both the x and y directions, so you may need to use vector addition to get the final result. I hope this helps!

 

[m2003]

Based on the given information, the electric field at location A can be calculated using the principle of superposition. This means that the total electric field at A is the vector sum of the electric fields produced by each dipole separately.

First, we can calculate the electric field produced by the dipole on the left. Using the dipole axis equation, we can find the electric field in the x-direction as E_x = [1/(4*pi*E)] * [q*s] * [1/r^3] * cos(theta), where theta is the angle between the dipole axis and the x-axis. In this case, theta is 90 degrees, so cos(theta) = 0. Therefore, the electric field in the x-direction is 0 N/C.

Next, we can calculate the electric field in the y-direction using the perpendicular axis equation. This gives us E_y = [1/(4*pi*E)] * [q*s] * [1/r^3] * sin(theta), where theta is the angle between the dipole axis and the y-axis. In this case, theta is also 90 degrees, so sin(theta) = 1. Plugging in the given values for q, s, and d, we get E_y = (9E9) * (3E-9 C * 0.001 m) * (1/(0.07 m)^3) * 1 = 77.14 N/C.

Now, we can calculate the electric field produced by the dipole on the right. Using the same equations as above, we get E_x = 0 N/C and E_y = -77.14 N/C. Note that the negative sign is due to the opposite direction of the electric field produced by this dipole compared to the first one.

Finally, we can add the electric fields from both dipoles together to get the total electric field at location A. Since the electric fields are in the same direction in the y-axis, the total electric field in the y-direction is simply the sum of the two, giving us E_y = 77.14 N/C. The total electric field in the x-direction is 0 N/C, since the two electric fields cancel each other out.

Therefore, the final answer for the electric field at location A is <0, 77.14, 0> N/C. This is because the electric field is only present in the y