How can the amount of work required to move two charges be calculated?

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To calculate the work required to move two charges, one must determine the initial and final electric potentials using the formula V=k(r)(q1+q2). The work done by an outside agent is related to the change in potential energy, which can be expressed as W=-qE*delta r. The discussion highlights that the potential V calculated represents the electrostatic potential at a specific point, not directly applicable to the problem at hand. It emphasizes the importance of analyzing the overall change in potential energy of the system rather than focusing solely on the electric potential at a point. Understanding these concepts is crucial for solving the problem effectively.
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Homework Statement


I get depressed and frustrated with every physics problem. The more studying and reading I do, the worse my grade gets. Talk about diminishing returns. Anyway, charges q1=1.65x10^-5 C and q2=-5.65x10^-5 C are placed 0.6 m apart. How much work must be done by an outside agent to move these charges slowly and steadily until they are 0.355 m apart?


Homework Equations



V=k/r(q1+q2)
deltaV= -E*delta r
-W=-qE*delta r

The Attempt at a Solution


Attempt at the solution is more than likely wrong
Vinitial=k/0.6(q1+q2)
Vfinal=k/0.355(q1+q2)
deltaV= final-initial

deltaV=-W*(-q)
W=deltaV/q
I got a really huge number. I used q=1.65x10^-5 because the positive charge will move to the negative charge.
 
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What does V = k/r(q1+q2) represent? It is the electrostatic potential due to what? Are the charges in the denominator as the equation seems to imply?
 
V=k*(q1+q2)/r is the equation to find the electric potential.
 
tooold said:
V=k*(q1+q2)/r is the equation to find the electric potential.
That unfortunately gives the electric potential at a point that is equidistant from both charges q1 and q2 at a distance of r, and will not aid you at all in this question.
You should consider how the overall potential energy of the system has changed.
 
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