Find the y-compnent of the force

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The discussion focuses on calculating the y-component of the force exerted on object B by object A, both of which are charged. The formula used is F = K(q1)(q2)/(r^2), where the calculated force is approximately 0.0047 N. A participant suggests that their calculation yields a similar result of 4.8 x 10^-3 N in the positive y direction. There is some confusion about why the problem specifically asks for the y-component when both objects are aligned along the y-axis. Overall, the conversation highlights the importance of careful reading in physics problems.
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Homework Statement


Object A, which has been charged to + 21.2nC , is at the origin. Object B, which has been charged to + 19.7nC , is at (x,y)= (0cm, 2.80cm) .

What is the y-component of the force (F a on b)y on B due to A

Homework Equations



F=K(q1)(q2)/(r^2)

The Attempt at a Solution



F= (9*10^9)*(21.2*10^-9)*(19.7*10^-9)/(.028^2)
= .0047N

Where am I messing up? Thanks
 
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kdizzle711 said:

Homework Statement


Object A, which has been charged to + 21.2nC , is at the origin. Object B, which has been charged to + 19.7nC , is at (x,y)= (0cm, 2.80cm) .

What is the y-component of the force (F a on b)y on B due to A

Homework Equations



F=K(q1)(q2)/(r^2)

The Attempt at a Solution



F= (9*10^9)*(21.2*10^-9)*(19.7*10^-9)/(.028^2)
= .0047N

Where am I messing up? Thanks

From your statement of the problem I get 4.8*10-3 N in the positive y direction.

Which is seemingly about the same as yours. Is there a preference for how the answer should be entered perhaps?
 
EDIT: oops! misread it. :redface:

Thanks, LowlyPion! :smile:

(though why does the question bother to ask for the y-component if everything's on the y-axis anyway? :rolleyes:)
 
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Edit: Agreed. The wording of the OP seems more designed to encourage careful reading than grasping the concepts. I suppose it is preparation for latter problem statements.
 
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