Question

Two identical +5.5 μC point charges are initially spaced 6.5 cm from each other. Part A If they are released at the same instant from rest, how fast will they be moving when they are very far away from each other? Assume they have identical masses of 1.0 mg .

Answer 1

v=2.044
`x10^(3)(m)/(s)`

Step-by-step explanation:

The initial total potential energy

Ep=
`(q1*q2)/(4\pi*E_(o)*d)`

`q1=q2=5.5uC=5.5x10^(-6)C\\ E_(o)=8.85x10^(-12)\\ d=6.5cm*(1m)/(100cm)=0.065m\\ Ep=(5.5x10^(-6)C*5.5x10^(-6)C)/(4\pi*8.85x10^(-12)*0.065m)\\ Ep=4.18 J`

The potential energy is converted to kinetic energy, each particle will have half of the potential energy so:

`m=1mg(1g)/(1000mg)*(1kg)/(1000g)=1x10^(-6)kg`

`Ek=(4.184J)/(2)=2.09J`

`Ek=(1)/(2)*m*v^(2) \\v^(2)=(Ek*2)/(m) \\v=\sqrt{(Ek*2)/(m)} \\v=\sqrt{(2.09J*2)/(1x10^(-6) kg)}`

v= 2.04x10³ m/s