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

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The work done by an external force in moving a unit positive charge from point A to point B is calculated using the formula W = kq ((1/rB) - (1/rA)), where k is the electrostatic constant. The discrepancy arises from a sign error in the initial derivation, which should reflect the change in potential energy. The work done corresponds to the change in energy, expressed as W = ΔE = E_final - E_initial. The key takeaway is that the work done is path-independent and solely depends on the initial and final positions of the charge. Clarifying the sign in the formula resolves the confusion regarding the work done.
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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