Question

Consider three charges 9, 92, 9, each equal to 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?




plz don't write useless things​

Answer 1

Answer:

See let the 9, 92 , and 9 be q1, q2, and q 3

Solution :

In a equilateral triangle, all sides are equal and is 600. Similarly, all sides are equal and let it be L.

In the figure, from ∆ BOD, Angle OBD = 30° and BD = L/2

AD = AC cos 300 = AC (√3/2) = L (√3/2)

For a centroid, AO from the point A = 2/3 of AD as the centroid divides the line in the ratio 2:1

Hence, 2/3 AD = (2/3) L (√3/2)

                      = L (1 / √3)

Force F1 on Q due to charge q at A is

= (1/4∏є0 )  ×Qq/(L2/3) = (3/4∏є0 )  × Qq / L2 along AO

Force F2 on Q due to charge q at B is

= (1/4∏є0 )  × Qq/(L2/3) = (3/4∏є0 )  ×Qq / L2 along BO

Force F3 on Q due to charge q at C is

= (1/4∏є0 )  ×Qq/(L2/3) = (3/4∏є0 )  × Qq / L2 along CO

Resultant of force F2 and force F3 by vector parallelogram of addition is

= (3/4∏є0 )  × Qq / L2 along OA

Total force on Q due to F1 and the resultant of the force F2 and F3 is

= (3/4∏є0 )  × Qq / L2 along AO +  (3/4∏є0 )  × Qq / L2 along OA

= 0 as vector magnitude is same and direction is opposite