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

[Ashu2912]

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!

 

[zhermes]

Settle down tiger :p
The work done, is the change in energy
[tex]
W = \Delta E = E_{final} - E_{initial}
[/tex]
You're just off by a minus sign.

 

[Ashu2912]

Can you get me the derivation?

 

[rock.freak667]

You already did the derivation, you are just off by a minus sign.

 

[sardar]

First of all, it is important to note that the work done in moving a unit positive charge in space by an external force is independent of the path taken. This is because the Coulumbic force, which is the force between two charged particles, is a conservative force. This means that the work done by this force is only dependent on the initial and final positions of the charge and not on the path taken to get there.

In this scenario, the work done by an external force equal and opposite to the Coulumbic force, in moving a unit positive charge from A to B, will be the same regardless of the path taken. This is because the external force is counteracting the Coulumbic force, and thus the net force on the charge is zero. Therefore, the work done by the external force is also zero.

Now, regarding the discrepancy between the two equations for work that you have derived, it is important to note that the electrostatic constant, k, is a positive value. This means that the work done by the Coulumbic force is always positive, as it is a force of attraction between two particles with opposite charges. Therefore, the work done by the external force must be equal in magnitude but opposite in direction, resulting in a negative value for work done.

In the equation W = kq ((1/rA) - (1/rB)), the negative sign is already included in the value of k, while in the equation given by your book, the negative sign is explicitly written before the value of k. Therefore, both equations are correct and will give the same result for the work done by the external force in moving a unit positive charge from A to B. It is just a matter of how the negative sign is represented in the equation.

I hope this explanation helps clarify the discrepancy and reinforces the concept that the work done in moving a unit positive charge in space by an external force is independent of the path taken.