Question

dipole moment of a system of charge +q distributed uniformly on an arc of radius R subtends an angle pi/2 at its centre where another charge -q is placed is

Answer 1

@nincompoop

Answer 2

@johnweldon1993

Answer 3

i can help

Answer 4

nevermind

Answer 5

@brett.dodd16 help needed

Answer 6

@confluxepic

Answer 7

@hartnn

Answer 8

@King.Void.

Answer 9

i an help you do this. we need to break the arc into small elements and do an integration. because its symmetrical, it should be OK, it will just be distance along axis from the point charge

would be helpful if you could post whatever formula you have been given for dipole moment so it looks the part. vector eqn, scalar eqn, little summation thingy with a big sigma....whatever.

incidentally you haven't said where the other charge is placed.....least i couldn't see that info

Answer 10

this is what i think you are describing. i have added the x and y ref and oriented it for ease of explanation. note the symmetry about the x axis that makes this workable, hence the chosen orientation

in terms of formulating the dipole moment, this is a simple way:\[\vec p = \Sigma q_i \ \vec r_i = \Sigma q_i \ x_{ i} + \Sigma q_i \ y_{ i} = \Sigma q_i \ x_{ i} + 0\]

IOW this is analogous to finding the centre of mass of the arc of a circle. by placing the -q at the origin, we can take the moment about that point of all the charge on the arc and replace the arc with a single equivalent charge +q at a point that should correspond with an analogous CoM calculation.