Calculating Electric Field due to a Dipole

  • Thread starter guitarman
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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 = 4 nC, s = 1 mm, and d = 6 cm, what is the electric field at location A?

As I am unaware of how to attach a file, here is a diagram
************+
************-
************|
************|
************|
************|
-***********|
+--------------A

the *s are there for formatting, or the else lines do not align

Homework Equations


(1/4*pi*epsilon)*(2qs/d^3) is the electric field of a dipole, on the dipole axis
and
(1/4*pi*epsilon)*(qs/d^3) is the electric field of a dipole, perpendicular to the dipole axis


The Attempt at a Solution


Plugging the values given, I used the latter equation for the x component, and by doing (9e9)(4e-9*1e-3)/(6e-2)^3 I obtained 166 N/C.

I used the first equation for the y component and obtained 333 N/C

I know there is no z component, so I input <166, 333, 0> N/C, yet this is wrong. Can somebody explain to me what I am doing? I attempted submitting -166 for the x component in case I had somehow misinterpreted the direction of the field, but that was also wrong. Does anybody understand what I am doing wrong?
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
In the dipole above A, the negative charge is closer to A . So the attractive force is more than the repulsive force. So the electric field is towards the dipole.
In the equatorial position the direction of the field is parallel to the dipole from positive charge to negative charge.
Hence the electric field due to two dipoles are in the same direction. So the net field is the sum of them.
 

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