drawing an fbd and doing force balance we get
vertical force balance
T*cos() = W {W is the
weight}
horizontal force balance
F= T*sin()
and also
F= kq2/r2 = 9*109*(10*10-9)2/ (0.005)2
F = 0.036 N
dividing both equations
tan() = F/W
W = F/(tan())
putting the values
W=0.2041 N
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