JEE Class main Answered
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Let OAB is the equilateral triangle of side 5 cm . Let each vertex of triangle has charge 2 μC .
Vertex O is at origin and side OA be on x-axis.
Force FA is electrostatic force on charge at O due to charge at A . Force FB is electrostatic force on charge at O due to charge at B .
Electrostatic force F between charges q1 and q2 that are separated by a distance d is
where K = 9 × 109 N m2 C-2 is Coulomb's constant
Magnitudes of forces FA and FB are equal because distance between charges are equal.
we get, |FA | = | FB | = 14.4 N
Net force on charge at O is
Magnitude of force is
Hence each charge at the vertex of triangle experiences a force 14.4 N due to other charges