Electric Dipole Moment and electric field

AI Thread Summary
The discussion centers on calculating the electric field at location A due to two dipoles with given parameters. The first dipole's contribution to the electric field in the x direction is determined to be zero, while the second dipole's contribution is calculated using the formula E(axis) = 2qks/d^3, yielding a value of 500. However, a mistake is identified in the calculation of the y component from the x-axis dipole, which is not zero as initially thought. The corrected y component is found to be 250, leading to a total electric field of <0,750,0>. The final answer reflects the combined contributions from both dipoles.
DeadFishFactory
Messages
25
Reaction score
0

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 = 6 nC, s = 1 mm, and d = 6 cm, what is the electric field at location A?

14-074-two_dipoles_A.jpg


Homework Equations



E(axis) = qk x (r-s/2)^2 - (r+s/2)^2)^-1

E(axis) = 2qks/d^3 when d>>s

k=8.99E9 N^2 m / C^2

The Attempt at a Solution



Since d>>s, the x-axis dipole's angle can be considered linear, and thus, the E-field for the x component is 0.

For the y-axis dipole, d>>s allows us to use the second equation, E(axis) = 2qks/d^3.

E = 2 (8.99E9) (6E-9) (0.001) / (0.06^3) = 500.

My answer would be <0,500,0>. I got the x and z components correct, but the y component is not. Why is that? Did I get the x component right for the wrong reasons or something?
 
Physics news on Phys.org
The y component of the electric field arising from the x-axis dipole is not 0.

ehild
 
Oh, whoops. You're right.

With the Y component for the x-axis, it would be E(y) = kqs/y^3 = (8.99E9 * 6E-9 * 0.001) / (0.06^3) = 250.

250 + 500 = 750; <0,750,0>

Thanks.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top