two point charges lie on the x axis. A charge of 6.2 is at the origin, and a cha

Question

two point charges lie on the x axis. A charge of 6.2 is at the origin, and a charge of - 9.2 is at x = 10 cm. What is the net electric field at x= -4 cm and x= 4 cm. A diagram showing distances (r) will be appreciated.

Explanation / Answer

At (A) i.e.x = -4.0 cm:- Put a test charge +q at given point =>Let F1 is the force produced on it by +6.2 uC:- =>F1 = kq1q2/r^2 =>F1 = 9 x 10^9 x 6.2 x q x 10^-3/(4 x 10^-2)^2 =>F1 = 13.95q x 10^10 N [direction -Xaxis,Let take it -ve] =>F1 = - 13.95q x 10^10 N =>Let F2 is the force produced on it by -9.5 uC:- =>F1 = kq1q2/r^2 =>F1 = 9 x 10^9 x 9.5 x q x10^-3/(14 x 10^-2)^2 =>F1 = 6.11q x 10^10 N [direction +Xaxis,Thus take it +ve] =>F1 = + 6.11q x 10^10 N =>F(net) = F1 + F2 =>F(net) = - 13.95q x 10^10 + 6.11q x 10^10 =>F(net) = -7.84q x 10^10 N =>E = F(net)/q = -7.84 x 10^10 N/C [-ve indicating the direction is in -Xaxis] At (B) i.e. x = +4.0 cm:- Put a test charge +q at given point =>Let F1 is the force produced on it by +6.2 uC:- =>F1 = kq1q2/r^2 =>F1 = 9 x 10^9 x 6.2 x q x 10^-3/(4 x 10^-2)^2 =>F1 = 13.95q x 10^10 N [direction +Xaxis,Let take it +ve] =>F1 = + 13.95q x 10^10 N =>Let F2 is the force produced on it by -9.5 uC:- =>F1 = kq1q2/r^2 =>F1 = 9 x 10^9 x 9.5 x q x10^-3/(6 x 10^-2)^2 =>F1 = 14.25q x 10^10 N [direction +Xaxis,Thus take it +ve] =>F1 = + 14.25q x 10^10 N =>F(net) = F1 + F2 =>F(net) = + 13.95q x 10^10 + 14.25q x 10^10 =>F(net) = +28.20q x 10^10 N =>E = F(net)/q = +28.20 x 10^10 N/C [+ve indicating the direction is in +Xaxis]

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