Help with two quick electric potential and voltage problesm

AI Thread Summary
To calculate the work needed to move a -6.0 µC charge to a point with a potential of +7.00 V, the correct formula is w = q * v, resulting in 0.000042 J, with the sign indicating work done by the field. For the second problem, the electron's speed when far away from a -0.150 µC charge can be determined using the equation W = K * (qQ/r), but the charge value used was incorrect, affecting the final velocity calculation. The correct charge value should be -0.150 µC instead of -0.125 µC. The final velocity should not include a negative sign as it pertains to magnitude only. Accurate calculations are crucial for both problems to ensure correct answers.
jhcollaz
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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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