Question

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 charge Q (with the same sign as q) placed at the centroid of the triangle????

Answer 1

Heyya mate!!!

Using symmetry here we see that that all froces

=) F1 = F2 = F3 = kQq/r^2

here r = AO = BO = CO

using trigonometry we can find " r "

Here we can see that length of each side of ∆ ABC is " L "

=) In ∆ BOD

Angle OBD = 30°

and BD = L/2

COS 30 ° = L/2r =) r = L/2 × sec30°

=). r = L/2 × 2/√3 = ( 1/√3 ) L -----(1)

so we can see tha all forces are equal

F1 = F2 = F3 = kQq/ (L/√3)^2 = 3kQq/L^2

and angle between each forces is 120°

using Lami's theorem if all forces are equal in magnitude and angle between then is also equal and planar then resultant force is 0

=) Fnet = F1 +F2 + F3 = 0

=) here we add forces using vectors to get net force = 0

so Net force on charge " Q " placed at centroid is zero.

Hope it helpful

Answer 2

Answer:

Explanation:

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 = ( $\sqrt{3}$/2 ) l and the distance AO of the centroid O

from A is (2/3) AD = (1/$\sqrt{3}$ ) l. By symmatry AO = BO = CO.

Thus,

Force F1  on Q due to charge q at A = $\frac{3}{4\pi E0} \frac{Qq}{l^{2} }$  along AO

Force F2 on Q due to charge q at B = $\frac{3}{4\pi E0} \frac{Qq}{l^{2} }$ along BO

Force F3 on Q due to charge q at C = $\frac{3}{4\pi E0} \frac{Qq}{l^{2} }$  along CO

The resultant of forces F2  and F3  is $\frac{3}{4\pi E0} \frac{Qq}{l^{2} }$  along OA, by the

parallelogram law. Therefore, the total force on Q =  $\frac{3}{4\pi E0} \frac{Qq}{l^{2} } (r-r)$=0  where r 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.