# Formula for Work done

Is the formula for workdone, - charge (Q) x electric potential (V) ? OR just Q x V ?
in my textbook it's Qx V (it doesn't have the -ve sign) but i saw in this video (from organic chemistry)
https://youtu.be/LzKMQByFSLc
he writes, -q x v
so which one is the formula?
Original post by Aleksander Krol
Is the formula for workdone, - charge (Q) x electric potential (V) ? OR just Q x V ?
in my textbook it's Qx V (it doesn't have the -ve sign) but i saw in this video (from organic chemistry)
https://youtu.be/LzKMQByFSLc
he writes, -q x v
so which one is the formula?

Work is done when you move a charge between 2 points in an electric field.
It depends on
1) if the charge is positive or negative
2) in which direction you move it - and thus the difference in potential (V) between those 2 points.
Original post by Aleksander Krol
Is the formula for workdone, - charge (Q) x electric potential (V) ? OR just Q x V ?
in my textbook it's Qx V (it doesn't have the -ve sign) but i saw in this video (from organic chemistry)
https://youtu.be/LzKMQByFSLc
he writes, -q x v
so which one is the formula?

Is the formula for workdone, - charge (Q) x electric potential (V) ? OR just Q x V ?

IMO, it is neither.
In physics, the product of electric potential and charge is the “electric potential energy of the charge” NOT work done.

As the change of electric potential energy is the product of charge and potential difference, and work done by the conservative force is equal to the negative of the change in electric potential energy (we can take it as by definition at A level), so

$W = ‒\Delta U$

$W = ‒q \Delta V$

The electric force is a conservative force. [A force for which the work done on a particle as it moves from an initial to a final position is independent of the path followed is called a conservative force.]

https://youtu.be/LzKMQByFSLc
The Youtube video that you have linked is relating work done with potential difference NOT work done with potential. Ensure that you see the difference.

The negative sign is important as it would tell us how we define the work done. Stonebridge has summarised the detail in post #2, I would “expound” the points below.

From the diagram above, note that negative and positive charges respond “differently” to electric field but the result is the same. Placing a stationary positive charge in a uniform electric field (ignore gravity, air resistance, etc), the electric field exerts a force on the positive charge in direction of the electric field, so work done by the electric field is positive, in moving the positive charge from high potential to low potential and there is a decrease in “electric potential energy of the charge”.
Apply the same analysis to the stationary negative charge in a uniform electric field, the electric field exerts a force on the negative charge in the opposite direction of the electric field. One may think that the work done by the electric field is negative but it is NOT. If we let go of the stationary negative charge, the negative charge would “want” to go to high potential ie the positive terminal as shown on the above diagram (right side). Since the direction of electric force and the direction of displacement, work done by the electric field is positive again, in moving the negative charge from low potential to high potential. From the viewpoint of a negative charge, in moving from low potential to high potential, there is a decrease in the “electric potential energy of the charge”. This could be hard to swallow for the first-timer.
Recall that electric potential energy is
$U = qV$
If the charge q is negative and the potential V is high, the electric potential energy will be “very negative” and this is why in moving a negative charge from low potential to high potential, there is a decrease in “electric potential energy of the charge”.
Original post by Eimmanuel
IMO, it is neither.
In physics, the product of electric potential and charge is the “electric potential energy of the charge” NOT work done.

As the change of electric potential energy is the product of charge and potential difference, and work done by the conservative force is equal to the negative of the change in electric potential energy (we can take it as by definition at A level), so

$W = ‒\Delta U$

$W = ‒q \Delta V$

The electric force is a conservative force. [A force for which the work done on a particle as it moves from an initial to a final position is independent of the path followed is called a conservative force.]

https://youtu.be/LzKMQByFSLc
The Youtube video that you have linked is relating work done with potential difference NOT work done with potential. Ensure that you see the difference.

The negative sign is important as it would tell us how we define the work done. Stonebridge has summarised the detail in post #2, I would “expound” the points below.

From the diagram above, note that negative and positive charges respond “differently” to electric field but the result is the same. Placing a stationary positive charge in a uniform electric field (ignore gravity, air resistance, etc), the electric field exerts a force on the positive charge in direction of the electric field, so work done by the electric field is positive, in moving the positive charge from high potential to low potential and there is a decrease in “electric potential energy of the charge”.
Apply the same analysis to the stationary negative charge in a uniform electric field, the electric field exerts a force on the negative charge in the opposite direction of the electric field. One may think that the work done by the electric field is negative but it is NOT. If we let go of the stationary negative charge, the negative charge would “want” to go to high potential ie the positive terminal as shown on the above diagram (right side). Since the direction of electric force and the direction of displacement, work done by the electric field is positive again, in moving the negative charge from low potential to high potential. From the viewpoint of a negative charge, in moving from low potential to high potential, there is a decrease in the “electric potential energy of the charge”. This could be hard to swallow for the first-timer.
Recall that electric potential energy is
$U = qV$
If the charge q is negative and the potential V is high, the electric potential energy will be “very negative” and this is why in moving a negative charge from low potential to high potential, there is a decrease in “electric potential energy of the charge”.

Amazing post ^^^