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Problem: What is the magnitude of the force F on the 5.0 nC charge in the figure? What is the direction of the force F on the 5.0 nC charge in the figure?

FREE Expert Solution

From Coulomb's law:

F=kq1q2r2r^

The angle between 5.0 nC and 10nC is obtained by:

θ = tan-1(4.0/3.0) = 53.1°

Force on 5nC due to 10nC:

F1=(9×109)(5×10-9)(10×10-9)[(4.0×10-2)2+(3.0×10-2)2]2[cos(53.1°)(-i^)+sin(53.1°)(-j^)]

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What is the magnitude of the force F on the 5.0 nC charge in the figure? 

What is the direction of the force F on the 5.0 nC charge in the figure?

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