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

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Homework Help Overview

The discussion revolves around calculating the work required to move two electric charges, specifically q1=1.65x10^-5 C and q2=-5.65x10^-5 C, from a distance of 0.6 m to 0.355 m apart. The subject area is electrostatics, focusing on electric potential and work done in moving charges.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply equations related to electric potential and work but expresses uncertainty about their correctness. Some participants question the meaning of the equations and the implications of the variables involved, particularly regarding the representation of electric potential and the relevance to the problem at hand.

Discussion Status

Participants are exploring different interpretations of the equations related to electric potential. Some guidance has been offered regarding the need to consider changes in potential energy rather than solely relying on the electric potential at a point. There is no explicit consensus on the correct approach yet.

Contextual Notes

The original poster expresses frustration with their understanding and performance in physics, indicating a possible emotional barrier to problem-solving. There may be assumptions about the setup of the problem that are being questioned, particularly regarding the application of the equations provided.

tooold
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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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