1.

Three vectors placed at corners of equilateral triangle having charge 'Q' and side length 'L' .Find the total force on a charge 'q' placed at centroid of triangle

Answer»

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.


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