Help with two quick electric potential and voltage problesm

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SUMMARY

This discussion addresses two electric potential and voltage problems involving a -6.0 µC charge and an electron near a fixed point charge of -0.150 µC. For the first problem, the work done by the electric field to move the charge to a potential of +7.00 V is calculated using the formula w=q*v, resulting in 0.000042 J. The second problem involves calculating the speed of an electron starting from rest at a distance of 74.5 cm from the charge, using the equation W = K*(qQ/r) and the kinetic energy formula, leading to a final speed of approximately 2.30311e7 m/s. The discussion emphasizes the importance of using the correct charge values and considering the sign of the charge in calculations.

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Homework Statement



1) How much work is needed (done by the field) to move a -6.0 µC charge from ground to a point whose potential is +7.00 V higher?

2)An electron starts from rest 74.5 cm from a fixed point charge with Q = -0.150 µC. How fast will the electron be moving when it is very far away?

It says my answer is not right. I only have two attempts at these problems so I am not sure what to do. Do i include the sign of the charge?

Homework Equations



For (1)

w=q*v

For (2)

W = (1/2)*m*V^2 -(1/2)*m*V^2
W = K*(qQ/r)

The Attempt at a Solution



(1)
w=q*v
=(6*10^-6) * 7
= 0.000042 J

(2)
W = K*(qQ/r)
=(9*10^9)* ( ((1.6*10^19)*(0.125*10^-6)) / 0.745 )
W = 2.41611*10^-16

W = (1/2)*m*V^2 -(1/2)*m*V^2
2.41611*10^-16 = (1/2)*(9.11*10^-6)*V^2 -0
v = 2.30311e7
 
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I would include a minus sign on (1) because work is done by the potential as it attracts the negative charge to the positive plate.
In (2), you wrote .125 for the charge when it says .15 in the question, and it has thrown off your answer. No minus in (2) since directions are not involved.
 

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