Object A, which has been charged to +29.9nC, is at the origin. Object B, which h

Question

Object A, which has been charged to +29.9nC, is at the origin. Object B, which has been charged to            -19.0nC, is at (x,y)=(0.0cm, 2.40cm). (could you please show me how to do these calculations, your help will be greatly appreciated, Thank you in advance)

a)What is the x-component of the force (Fa on b)x on B due to A? (Fa on b)x=____N
b)What is the y-component of the force (Fa on b)y on B due to A? (Fa on b)y=____N
c)what is the x-component of the force (Fb on a)x on A due to B?                                                                 d)what is the y-component of the force (Fb on a)y on A due to B?

Explanation / Answer

Since the charges are opposite, the objects will be attracted towards each other.

A) Because the object at Y lies on the y-axis, there is no x-component of the attractive force. Hence, it is 0.

B) Use Couloumb's Law. FB,A = (8.99E9)(29.9E-9)(19E-9)/(.024^2) = .00887 N. Personally, I find taking out the parity of the charges (pos or neg) helpful and then determining the effects the charges have after the numbers have been determined. Because this is an attractive force and particle B has a more positive y-coordinate than does particle A, the force will point in the negative y direction. Hence, the force is -.00887 N.

C) 0. The origin lies on the y-axis.

D) This will be the opposite of B since particle A's y-coordinate is less positive. The force is .00887 N.

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