Numerical-3

Q.  Consider three charges q1 , q2 , q3 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, as shown in Fig. 1.9? 

Solution 

In the given equilateral triangle ABC of sides of length l, if

we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3/2 ) l and the distance AO of the centroid O from A is (2/3) AD = (1/ 3 ) l. By symmatry AO = BO = CO.

Thus, Force F1 on Q due to charge q at A = 3Qq / 4πεo l^2   along AO

 Force F2 on Q due to charge q at B = 3Qq / 4πεo l^2   along BO

 Force F3 on Q due to charge q at C = 3Qq / 4πεo l^2   along CO

The resultant of forces F2 and F3 is  3Qq / 4πεo l^2   along AO

by the parallelogram law. 

Therefore, the total force on Q = 3Qq / 4πεo l^2 (r^ - r^) = 0

where rˆ is the unit vector along OA

It is clear also by symmetry that the three forces will sum to zero.

Suppose that the resultant force was non-zero but in some direction.

Consider what would happen if the system was rotated through 60º

about O.

Numerical-4

Q. Consider the charges q, q, and –q placed at the vertices

of an equilateral triangle, as shown in Fig. 1.10. What is the force on

each charge? 

Solution 

The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in Fig. 1.10. 

By the parallelogram law, the total force F1 on the charge q at A is given by

F1 = F r^1

where r^1 is a unit vector along BC.

The force of attraction or repulsion for each pair of charges has the

same magnitude 

F = q^2 / 4πεo l^2 

The total force F2 on charge q at B is thus F2 = F r^2, where r^2 is a unit vector along AC. 

Similarly the total force on charge –q at C is F3 = 3 F n^

where n^ is the unit vector along the direction bisecting the ∠BCA.

It is interesting to see that the sum of the forces on the three charges is zero, i.e.,

F1 + F2 + F3 = 0