Note that your answer may be positive or negative, depending onthe direction of

Question

Note that your answer may be positive or negative, depending onthe direction of the force.

Explanation / Answer

You can use the Coulomb's law: And a plus is it's only in one dimension...which means we don'treally have to worry about vectors. It's either plus for positive xdirection or minus for negative x direction. Coulomb's Law is: F = kqQ/r2 --> k is the Coulomb's constant(8.988 x 109N•m2/C2), q isone charge, Q is the other, r is the distance betweenthem. For the charge at x=-1.695m: Force on q3 due to q1 F13= [(8.988 x109 N•m2/C2)(10.5x10-9C)(54.5x10-9C)]/(.555m)2= 1.66 x 10-5N --> this is the magnitude of theforce For the force on q3 due to q2: F23 = [(8.988 x109 N•m2/C2)(40.0 x10-9C)(54.5x10-9C)]/(1.140m)2 =1.51x10-5N --> this is the magnitude of the force think about this conceptually now that you know the forces due toeach charge on the one between them. the force between q1 and q3 is attractive, sothe force on q3 is in the negative x direction, towardthe negative charge, so the force is -1.66 x 10-5N. the force between q2 and q3 is repulsive, so the force acting on q3is away from q1, also in the negative direction, so the force on q3due to q1 is -1.51x10-5N. so add the forces together to get the net force. FT=-3.17x10-5N; it is negative because it isin the negative x direction. Please rate.

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