Three charges +2*10^_6,+3*10^-6and +4*10^_6 are placed at the corners of the equilateral triangle ABC of each side 0,2m
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The resultant force on charge Q because of two equal forces because of other two Q charges planed on equilateral triangle will be given as:-
Fnet = 2Fcos(ϴ/2) ---------------(1)
F=kQ1Q2/r2 --------(2)
in this question Q1=Q2 =Q3 =Q
Given
Q=2 *10-6 c
K=9*109 N-m2/c2
r= .05 m
ϴ =60 degree
By using equation 1 and 2 and given quantities
Fnet = 2(9*109 *(2 *10-6 )2/(.05)2 cos30
=24.94 N
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