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Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Potential Energy
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Set 1 Problem 4 states:
A stationary charge of 76 `microC is located at the origin, and a charge of 21 `microC is located at the point ( .32 m,0,0).
Estimate, based on the force at the initial and final points, the work required to move the second charge to the point ( .2112 m,0,0).
What is the approximate potential energy change associated with moving the second charge between these points?
Solution
We find the force exerted on the moveable charge at each of the two points, and use these forces to estimate the average force:
The force at the original point is found from Coulomb's Law to be (9 x 10^9 N m^2/C^2)( 76 x 10^-6 C)( 21 x 10^-6 C)/( .32)^2 = 140.2 N.
The force at the second point is found from Coulomb's Law to be (9 x 10^9 N m^2/C^2)( 76 x 10^-6 C)( 21 x 10^-6 C)/( .2112 m)^2 = 322 N.
The average of these forces is 231.1 N.
Assuming this to be the average force on the charge over the interval between the points (it is not in fact equal to the average; it is higher than the average but the average gives us a reasonable approximation in this case), we find the work using this force and the displacement from the first point to the second.
The displacement is ( .2112 m - .32 m) = -.1089 m.
The force that moves the charge from .2112 meters to .32 meters is equal and opposite to the force on the charge, or to +-231.1 N. This is the force that does the work.
So the work done is (-.1089 m)(-231.1 N) = 25.16 Joules.
The force is conservative, so this work, if positive, could be recovered in the form of kinetic energy if the particle was released. If negative, the work could have been performed at the expense of an equal amount of kinetic energy.
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To answer the question on potential energy change, would it be correct to just say the Potential Energy decreased by 25.16 Joules.
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The change in PE is positive. I'll clarify this in two ways, the first more intuitive, and the second in terms of the 'official' definition.
The calculation given in the solution was for the work that had to be done on the charge by the force that moves it against the field.
By analogy with raising an object against gravity, the work done against a conservative force is equal the change in potential energy.
In this case, since the electrostatic force was directed away from the origin, the force required to move the charge was directed toward the origin, and the displacement of the charge was toward the origin. So this force did positive work, increasing the PE of the system.
The 'official' definition of change in PE is that the change in PE is equal and opposite to the work done by the conservative force. In this case the conservative force is the electrostatic force, which is directed away from the origin. During the displacement, which is directed toward the origin, the conservative force and the displacement have opposite directions so the work done by the conservative force is negative. The change in PE, being equal and opposite to the work done by the conservative force, is therefore positive.
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