A point charge Q1 = +4.9 uC is fixed in space, while a point charge Q2 = +2.1 nC, with mass 6.6 ug, is free to move around nearby.  Part (a) If Q2 is released from rest at a point 0.36 m from Q1, what will be its speed, in meters per second, when it is 0.79 m from Q1?

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Chapter1: Units, Trigonometry. And Vectors
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A point charge Q1 = +4.9 uC is fixed in space, while a point charge Q2 = +2.1 nC, with mass 6.6 ug, is free to move around nearby. 

Part (a) If Q2 is released from rest at a point 0.36 m from Q1, what will be its speed, in meters per second, when it is 0.79 m from Q1?

Expert Solution
Step 1: Calculation with given data

Charge Q1= +4.9 uC = +4.9*10-6C

Charge Q2=+2.1 nC = +2.1*10-9 C

Mass = 6.6 ug = 6.6*10-9 kg


a) The initial potential energy changed to kinetic energy and potential energy at the new position.

That is,

U subscript i equals K E subscript f plus U subscript f
fraction numerator k Q subscript 1 Q subscript 2 over denominator r subscript 1 end fraction equals 1 half m v squared plus fraction numerator k Q subscript 1 Q subscript 2 over denominator r subscript 2 end fraction
1 half m v squared equals fraction numerator k Q subscript 1 Q subscript 2 over denominator r subscript 1 end fraction minus fraction numerator k Q subscript 1 Q subscript 2 over denominator r subscript 2 end fraction
1 half m v squared equals k Q subscript 1 Q subscript 2 open square brackets 1 over r subscript 1 minus 1 over r subscript 2 close square brackets
v equals square root of fraction numerator 2 k Q subscript 1 Q subscript 2 over denominator m end fraction open square brackets 1 over r subscript 1 minus 1 over r subscript 2 close square brackets end root

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