Consider three charges, #q_1=q_2=q_3=q#, at the vertices of an equilateral triangle of side #l#. What is the force on a charge #Q# (with the same sign as #q#) placed at the centroid of the triangle? Solve using parallelogram law of vectors.

Question

12146 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-16T18:07:08

The force is zero.

#sf(F=k.(q_1q_2)/(r^2))#

#sf(F)# is the force.

#sf(k )# is the Coulombic Constant.

#sf(q_1)# and #sf(q_2)# are the charges.

#sf(r)# is the separation.

An equilateral triangle looks like this:

keywordsuggest.org

#sf(h_a=h_b=h_c)#

Since all the charges are equal and the distance between them is the same the forces look like this:

MFDocs

From the symmetry of the situation you can see that the net force must be zero.

If you wish to show this you can set Q to be at the origin.

You can resolve #sf(F_1)# and #sf(F_3)# to give #sf(F_(res))#.

Because we have an equilateral triangle then #sf(theta=60^@)#.

#:.##sf(F_(res)=F_1costheta+F_3costheta)#

Since #sf(F_1=F_3=F)# then:

#sf(F_(res)=2Fcostheta=2Fxx0.5=F)#

This is acting upwards. It is exactly balanced by #sf(F_2)# acting downwards. Since #sf(F_2=F)# then the net force on Q will be #sf(F-F=0)#