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Three point charges q , -4q and 2q are placed at the vertices of an equilateral triangle ABC of side 'l' as shown in the Fig. Obtain the experssion for the magnitude of the resultant electric force acting on the charge q. |
Answer» Solution :As shown in Fig , force `vecF_(AC) = (1)/(4pi in_(0)) (q (2q))/(L^(2))` along `CA = (1)/(4pi in_(0)) * (2q^(2))/(l^(2))` along CA and `vecF_(AB) = (1)/(4pi in_(0)) * (q(4q))/(l^(2))` along AB `= (1)/(4pi in_(0)) * (4q^(2))/(l^(2))` along AB.  If the magnitude of the resultant electric force acting on the charge q placed at A be `F_(A)` , then `F_(A)= SQRT((F_(AC))^(2) + (F_(AB))^(2) + 2 F_(AC) . F_(AB) cos 120^(@))` `= sqrt(((2q^(2))/(4pi in_(0) l^(2)))^(2) + ((4q^(2))/(4 pi in_(0) l^2))^(2) + 2 ((2q^(2))/(4pi in_(0) l^(2)))^(2) ((4q^(2))/(4pi in_(0) l^(2))) (- (1)/(2)) ) vecF_(A)` `= (2q^(2))/(4 pi in_(0) l^(2)) sqrt(1 + 4 - 2) = (SQRT3 q^(2))/(2pi in_(0) l^(2))` |
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