NEET Class neet Answered
consider the charges q,q and -q placed at the vertices of an equilateral triangle of each side l. The sum of forces acting on each charge is
Asked by kapilakarthikeyan | 29 Jun, 2019, 06:53: PM
Expert Answer
Figure shows an equilateral triangle OAB , vertex O coincides with origin of x-y coordinate system and side OA along x-axis.
let us assume vertex O and A has the charge +q and vertex B has charge -q.
At vertex O, repulsive force F1 is along -ve x-direction, direction of attractive force F2 make angle 60° with x-axis as shown in figure
At vertex A, repulsive force F3 is along +ve x-direction, direction of attractive force F4 make angle 60° with x-axis as shown in figure
At vertex B, forces F5 and F6 are attractive , they act along the sides as shown in figure.
Magnitude |F| of all the forces is same , |F| = q2 / [ 4πεo l2 ] , where l is side length of triangle
Resultant force FO at origin = F1 + F2 = |F| { -i + (1/2)i + (√3/2)j } = (1/2) |F| [ - i + √3 j ]
Resultant force FA at vertex A = F3 + F4 = |F| { +i - (1/2)i + (√3/2)j } = (1/2) |F| [ i + √3 j ]
Resultant force FB at vertex B = F3 + F4 = -(2|F| cos 30) j = -√3 |F| j
From above equations, we get, sum of FO + FA + FB = 0
Answered by Thiyagarajan K | 30 Jun, 2019, 09:56: AM
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