1) Coulomb\'s law for the magnitude of the force F between two particles with ch

Question

1) Coulomb's law for the magnitude of the force F between two particles with charges Q and Q separated by a distance d is |F|=K|QQ|d2, where K=140, and 0=8.854×1012C2/(Nm2)is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -11.5 nC , is located at x1 = -1.735 m ; the second charge, q2 = 37.0 nC , is at the origin (x=0.0000).  

Part A:-  What is the force exerted by these two charges on a third charge q3 = 45.0 nC placed between q1 and q2 at x3 = -1.145 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Explanation / Answer

q1 = -11.5*10^-9 C

x1 = -1.735 m

q2 = 37*10^-9 C

x2 = 0

q3 = 45*10^-9 C

x3 = -1.145 m

force exerted by q1 on q3

F13x = -k*q1*q3/(x3-x1)^2

F13x = -(1/(4*pi*8.854*10^-12))*(11.5*10^-9*45*10^-9/(-1.145+1.735)^2)

F13x = -1.34*106-5 N

force exerted by q2 on q3

F23x = -k*q2*q3/(x3-x2)^2

F23x = -(1/(4*pi*8.854*10^-12))*(37*10^-9*45*10^-9/(-1.145-0)^2)

F23x = -1.14*10^-5 N

Force exerted F = F13x + F23x = -2.48*10^-5 N

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