Force and Torque on a dipole

  • Thread starter haroharo
  • Start date
  • #1
6
0

Homework Statement


A point charge Q1 is distance r from the center of a dipole consisting of charges +/- q2 separated by distance s. The charge is located in the plane that bisects the dipole. At this instant, assuming r>> s,

Part A: what is the magnitude of the force on the dipole?
Part B: what is the magnitude of the torque on the dipole?

In the space provided, enter the factor that multiplies (1/ε0) in your answer. Express this factor in terms of q2, Q1, s, π, and r.

Homework Equations


1. Edipole=(1/4πε0)(2p/r3)
2. p=qs
3. F=qE

4. τ=pEsinϑ

The Attempt at a Solution



Part A
Substituting equation 1and 2 into equation 3
F = qE
= (q2)(1/4π)(2q2s/r3)

Part B
τ = pEsinϑ
= (q2s)(1/4π)(2q2s/r3)sin90º
= (q2s)(1/4π)(2q2s/r3)


I got a feeling I'm missing an important step involving Q1 in both parts.
Help's much appreciated =)
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
Since r>>>>s force on +q and -q are equal and opposite. So what will be the net force on the dipole?
Torque = forceX distance.
Find the force between Q1 and q and calculate the torque.
 
  • #3
6
0
thanks for the reply rl.bhat
but for Part A, the question is looking for the magnitude therefore the force on one end of the dipole and not the net force. the net force on a dipole would be zero of course.

As for part b, i doubt that the solution can be that simple. The question must be solved through algebraic expressions only. Solving it in this manner would require an extensive amount of substitution.

If solved simply by using the forces between Q1 and +/-q2, we would have to add their adjacent component values requiring inverse tangent and cosine laws
 
  • #4
rl.bhat
Homework Helper
4,433
7
The expression which you have given for torque is for a dipole kept in a uniform electric field. Since r >>> s , you can assume that the field is uniform.
In your solution of part B, you have to substitute the field due to charge Q1 at q2.
 
  • #5
181
0

Homework Equations


1. Edipole=(1/4πε0)(2p/r3)
2. p=qs
3. F=qE
1 gives the radial dipole field at a point on the line of the dipole charges

At points on the "bisecting plane" the field due to the dipole is "tangential"
Edipole=(1/4πε0)(p/r3)
(Although the point of the problem is to derive this)
This field exerts a force on the point charge,
Q1(1/4πε0)(p/r3)

and the point charge exerts an equal and opposite force on the dipole.

Note the moment of this couple is equal and opposite to the torque,
necessary for conservation of angular momentum.
 
Last edited:
  • #6
6
0
if understand part b properly
t = force x distance
where force is the line action, therefore s sin ϑ which is basically s
force is the force between q2 and q1
so overall it would be
Q1(1/4πε0)(q2s/r3)s
 
  • #7
181
0
For ease of reading/writing, I'll call the point charge Q and the dipole charges+/- q.

The magnitude of the force exerted by Q on +/-q is F = Qq/(4 pi epsilon0 r^2)

a) Add the forces vectorially (triangle of forces) to find the resultant R.
Note that the triangle is similar to the one with sides r,r and s:
s/r = R/F

b) F*s
 
  • #8
6
0
awsome i got it
big thanks davieddy!
 

Related Threads on Force and Torque on a dipole

Replies
1
Views
11K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
4
Views
845
  • Last Post
Replies
11
Views
1K
Replies
1
Views
2K
Replies
1
Views
3K
Replies
3
Views
4K
Replies
4
Views
6K
  • Last Post
Replies
3
Views
2K
Top