EK 1001 Physics: Electricity and Magnetism

[greenseeking]

If q1 and q2 are both positively charged and q2 is released, what is the maximum velocity that can be achieved by q2? Please see figure attached.
Additional info: E is the electric field created by q1. V is voltage given point in the field E. Electric field created by Q2 is negligible compared to E. m1 is the mass of q1. m2 is mass of q2.

Answer: sqrt(2kq1q2/m2r)

Why isn't is sqrt(kq1q2/m2r)?
This is how I did it:
kq1q2/2r=1/2mv^2

Why is the "distance" r not 2r?

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[Romz]

If q1 and q2 are both positively charged and q2 is released, what is the maximum velocity that can be achieved by q2? Please see figure attached.
Additional info: E is the electric field created by q1. V is voltage given point in the field E. Electric field created by Q2 is negligible compared to E. m1 is the mass of q1. m2 is mass of q2.

Answer: sqrt(2kq1q2/m2r)

Why isn't is sqrt(kq1q2/m2r)?
This is how I did it:
kq1q2/2r=1/2mv^2

Why is the "distance" r not 2r?

First, Lmao at your diagram


What is colombs law: The force between two charges related by distance R. From what I can tell, the diagram is showing you that q2 is twice the distance from an arbitrary point r than is q1. Does this matter? Are we given scalar quantities which allow us to compute the TRUE value of r between q1 and q2? Nope. What your expression is saying is that we double the distance between q1 and q2 such that we get a diagram that shows we've got an r of 4r.

Tl: dr: The distance between the two points is r in terms of the equation and not 2r. You misused the equation.

Hope this clears things up.

 

[greenseeking]

lol. thanks.