Recent content by LabBioRat

Eddy Current Braking System First of all, how do eddy currents works? Second of all, how are they used as brakes?

Oh my goodness! Thank you for pointing out that fatal mistake!

That was after I summed up all of the charges and vectors. All of the charges evaluated to 0q, and the vectors from each of the charges all evaluate to ( 0\widehat{i} +d\sqrt{3}/2 \widehat{j} ) Thus my final answer would be 0q( 0\widehat{i} +d\sqrt{3}/2 \widehat{j}) I apologize for the...

Doesn't the -2q belong on the outside, just like the other charges? My origin is now at (0,0), and the position of the (-2q) charge is (0, d\sqrt{3}/2). I do not quite understand your warning.

(-2q)(0\widehat{i} + d\sqrt{3} /2\widehat{j})+ q( (-d/2)\widehat{i} +0\widehat{j} ) +q( (d/2)\widehat{i} + 0\widehat{j} ) = 0q( 0\widehat{i} +d\sqrt{3}/2 \widehat{j} )

(-2q)(0\widehat{i} + 0\widehat{j})+ q( (-d/2)\widehat{i} -d \sqrt{3} /2\widehat{j} ) +q( (d/2)\widehat{i} -d\sqrt{3} /2\widehat{j} ) = 2q( -d\sqrt{3}\widehat{j} )

Since the total charge of the configuration is 0, the dipole moment does not depend on a specific origin. I choose the origin to be at the (-2q) charge.

Gabbagabbahey, thank you for the warm welcome. I am treating dipole moment as a vector quantity in the Cartesian Coordinate system.

Homework Statement Find the value of the dipole moment of the distribution of charges. -2q on top of equilateral triangle +q on each of the other points. Each charge is separated by distance d. Homework Equations p = \sum (q_i)(r_i) The Attempt at a Solution p =...