Work done in moving a unit positive charge in space by an external force

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Discussion Overview

The discussion centers around the calculation of work done by an external force in moving a unit positive charge in an electric field, specifically addressing the differences in derived formulas and the interpretation of energy changes. The scope includes theoretical reasoning and mathematical derivation.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant derives the work done as W = kq ((1/rA) - (1/rB)), while another participant suggests the correct expression is W = kq ((1/rB) - (1/rA)), indicating a disagreement over the sign of the expression.
  • Another participant emphasizes that the work done is related to the change in energy, expressed as W = ΔE = E_{final} - E_{initial}, pointing out that the initial derivation is only off by a minus sign.
  • A request for the derivation of the work done is made, indicating interest in understanding the mathematical steps involved.

Areas of Agreement / Disagreement

Participants do not reach consensus on the correct expression for work done, with multiple competing views on the sign of the derived formula and its implications for energy change.

Contextual Notes

The discussion highlights potential confusion regarding the derivation of work done in an electric field and the interpretation of energy changes, but does not resolve the discrepancies in the formulas presented.

Ashu2912
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Consider a positive charge +q placed at the origin. Let A and b be two points in the space with position vectors rA and rB respectively. What will be the work done by an external force equal and opposite to the Coulumbic force, in moving a unit positive charge from A to B, irrespctive of the path AB?

I have derived it as W = kq ((1/rA) - (1/rB)), but my book tells that W = kq ((1/rB) - (1/rA)), where k = electrostatic constant. Please help me!
 
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Settle down tiger :p
The work done, is the change in energy
<br /> W = \Delta E = E_{final} - E_{initial}<br />
You're just off by a minus sign.
 
Can you get me the derivation?
 
You already did the derivation, you are just off by a minus sign.
 

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