force due to 0.8
F1 = k q1 / r1^2 = 9* 10^9* 0.8* 10^^-12 / ( 0.095^2)
F1 =0.798 N
angle made F1 with horizontal
x = arcsin ( 8.2 / 9.5) = 59.673
force due to 0.6
F2 = 9* 10^9* 0.6* 10^-12 / (0.095^2 - 0.082^2)
F2 = 2.347N
magnitude of net force
F^2 = F1^2 + F2^2 + 2 F1 F2 cos ( 180 - x)
F = 2.08 N
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