A charge 23.41 mC is placed to the left of another charge 90.23 mC on a plane, a

Question

A charge 23.41 mC is placed to the left of another charge 90.23 mC on a plane, as shown in the figure. Another charge of -9.051 ?C and mass 34.81 g (depicted as a blue sphere) is placed at rest at a distance 28.94 cm above the right-most charge and released. What is the speed of the -9.051 ?C charge if it reaches the gray position labeled A?

Image: http://i772.photobucket.com/albums/yy6/tantarero19/ScreenShot2013-09-04at54917PM_zps4fa4af7b.png

A charge 23.41 mC is placed to the left of another charge 90.23 mC on a plane, as shown in the figure. Another charge of -9.051 ?C and mass 34.81 g (depicted as a blue sphere) is placed at rest at a distance 28.94 cm above the right-most charge and released. What is the speed of the -9.051 ?C charge if it reaches the gray position labeled A?

Explanation / Answer

Solve with conservation of Energy

Initial PE = kqq/r + kqq/r

(9 X 10^9)(-9.051 X 10^-6)(90.23 X 10^-3)/(.2894) + (9 X 10^9)(-9.051 X 10^-6)(23.41 X 10^-3)/(.4326)

PE = -29806 J

Final PE = (9 X 10^9)(-9.051 X 10^-6)(90.23 X 10^-3)/(.22505) + (9 X 10^9)(-9.051 X 10^-6)(23.41 X 10^-3)/(.09645) = -52431 J

Now...

-29806 = -52431 + .5mv^2

22625 = (.5)(.03481)(v^2)

v = 1140 m/s

v =

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