Charges `5.0 xx 10^(-7) C , -2.5 xx 10^(-7) Cand 1.0 xx 10^(-7) C` are
held fixed at the three corners A, B, C of equilateral triangle of side 10cm respectively. Find the electric force on the charge at C due to the rest two.

Explanation

The force on C due to A
`=(1)/(4piepsilon_0) ((5 xx 10^(-7) C ) ( 1xx 10 ^(-7) C))/(0.05m)^2)`
`=9xx 10^9 Nm^2 C ^(-2) xx (5xx10^(-14)C^2)/(25 xx 10 ^(-4) M ^2) =0.18 N`.
This force acts along AC. The force on C due to B
`= (1)/(`=1/(4^piepsilon_0) ((2.5 xx 10 ^(-7) C) (1 xx 10^(-7) C) )/((0.05m)^2) =0.09 N`.
This attractive force acts along CB. As sthe triangle is
equilateral, the angle between these two forces in `120^@`
The resultant electric force on C is
`[(0.18 N) ^ 2 = + (0.09 N) ^ 2 (0.18 N) (0.09 N) (cos120^@)] ^(1//2)`
`= 0.16 N.`
The angle made by this resultant with CB is
`tan ^(-1) (0.18 sin 120^@)/ (0.09 +0.18 cos 120^@) = 90^@.`