Two charges of 1×10^-9 C, each are 8 cm apart in air. Find the magnitude and direction of the force exerted by the charges on a third charge of 5.0 ×10^-11 C that is 5 cm distant from each of the first two charges.

D ^ E | F \ | / \ | / S_ _ _ _ C _ _ _ _P A B Suppose the charges at A and B are +ve and the charge at C is +veAngle DCE= Angle DCF= 45° (half of 90°) A exerts repulsive force CF on C and B exerts repulsive force CE on C. CF=CE (1) because A and B are equal in magnitude and nature of charge. Breaking CF and CE into their components CD and CP for charge AAnd CD and CS for charge B, we getCP = CF sin 45CS = CE sin 45 or -CF sin 45 (from 1, - ve due to opposite direction) Adding CP and CS, we get that the horizontal component of force is zero. CD = CF cos 45 + CE cos 45=> CD = 2* CF cos 45 (from 1)=> CD = 2× (kAC/(5)(5)) × cos 45= 2× (9×10^9×1×10^-9×5×10^-11/ 25) × cos 45=2.54×10^-11C, directed vertically upwards, along the CD vector.

I made a diagram but the it didn`t appear as I had expected. If you can draw a force diagram by yourself, that would be nice.