three charges of 5micro coloumb are placed at the vertices of equilateral triangle of side 10cm the force on 1micro coulomb of charge at centre of triangle will be
Asked by pariharhimani454 | 14th Apr, 2021, 12:19: PM
Figure shows the system of charge configuration on vertex of equilateral triangle. Direction of forces also indicated in figure.
Since the given equilateral triangle has 10 cm side, distance between charge at centre and charge at any vertex is 5 √3 cm .
All forces have equal magnitude
| F1 | = | F2 | = | F3 | = K × ( 5 × 10-12 ) / ( 75 × 10-4 ) = ( 9 × 109 × 5 × 10-12 ) / ( 75 × 10-4 ) = 0.06 N
where K = 1/( 4πεo ) = 9 × 109 N m2 C-2 is Coulomb's constnat
Resultant of F2 and F3 is 2F cos60 N along the direction OA , where F = 0.06 N
Resultant of F2 and F3 = FR = 0.06 N
Since FR and F1 are equal in magnitude but opposite in direction, net force is zero.
Sum of the forces acting at charge at centre is zero.
Answered by Thiyagarajan K | 14th Apr, 2021, 01:34: PM
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