B Why there is a negative sign in the formula of calculating work done?

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$$W = - \int _ { a } ^ { b } \vec { F } \cdot d \vec { r }$$
( The Force here is referring to the applied force, When moving a positive charge towards another positive charge(stationary) / field

In this formula why there is a negative sign in the formula? I am not asking the sign of the total workdone at the end of the calculation, but the sign in the formula that is before calculation. from where the sign comes from?does it have anything to do with the co-ordinate system,if yes then can the sign be changed if we take a different quadrant?.
 
Mmm... where did you get that equation?
 
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Mmm... where did you get that equation?
It is in most of the physics books like The Feynman lectures on physics vol. 2,Electricity and Magnetism by Edward M Purcell etc..
 
$$W = - \int _ { a } ^ { b } \vec { F } \cdot d \vec { r }$$
Ok, so this equation is the work done by what force? And from what point to what point?
 

Doc Al

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$$W = - \int _ { a } ^ { b } \vec { F } \cdot d \vec { r }$$
( The Force here is referring to the applied force, When moving a positive charge towards another positive charge(stationary) / field

In this formula why there is a negative sign in the formula? I am not asking the sign of the total workdone at the end of the calculation, but the sign in the formula that is before calculation. from where the sign comes from?does it have anything to do with the co-ordinate system,if yes then can the sign be changed if we take a different quadrant?.
In Feynman's version, that F refers to the electric force on the particle, not the applied force. The applied force is the negative of that, thus the minus sign.

(Feynman is calculating the work done against the electrical forces: the work done by the applied force.)
 
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from where the sign comes from?
The sign comes from the definition of what force is F and which system is doing the work. Just pay attention to which force is being calculated and which object is doing the work and then usually the sign should be obvious. Sometimes the F is from a different source than the W is calculated for.
 
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Ok, so this equation is the work done by what force? And from what point to what point?
Work done by the applied force from point b to a
 
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collinsmark

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It also may involve who or what is doing the work, or on whom or on what the work is being applied.

For example if you slowly push a positive charge toward another positive charge, then the work you do is positive since you are doing the work. This is because the force that you are applying is in the same direction as the displacement.

On the other hand, from the perspective of the positive charge, the work that it does on you is negative, since its equal and opposite force is in the opposite direction of the displacement.

[Edit: similarly, if you were to hold onto a positive charge and slowly move away from another positive charge, the work that you do is negative. The force that you are applying is in the opposite direction of displacement.]
 
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In Feynman's version, that F refers to the electric force on the particle, not the applied force. The applied force is the negative of that, thus the minus sign.

(Feynman is calculating the work done against the electrical forces: the work done by the applied force.)
But why the applied force has to be negative? I know it is equal and opposite, but, for example, consider the source charge is at the origin of a 2d coordinate system and the other positive charge is coming from the q3 quadrant (where both x and y have negative values), then in this case the electrical force would be negative and the applied force would be positive. So does the equation depend upon the coordinate system?
 

Doc Al

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But why the applied force has to be negative?
The applied force is opposite to the electrical force, which is ##\vec{F}##. So the applied force equals ##-\vec{F}##. (That doesn't mean it's 'negative' -- that depends on your coordinate system.)
 
But why the applied force has to be negative? I know it is equal and opposite, but, for example, consider the source charge is at the origin of a 2d coordinate system and the other positive charge is coming from the q3 quadrant (where both x and y have negative values), then in this case the electrical force would be negative and the applied force would be positive. So does the equation depend upon the coordinate system?
The work done by a force from a point ##a## to a point ##b## (if the force is conservative) is, by definition $$\int_{a}^{b}\vec{F}\cdot\text{d}\vec{r}$$ (where ##\vec{F}## refers to the applied force. So, the force you need to apply is equal and opposite to the force that is doing the other particle, then the applied force is ##-\vec{F_{e}}## (now ##\vec{F_e}## is the force done by the electric charge) and you have the - you want.

It doesn't make sense to talk about "positive" and "negative" forces, forces are vectors, not numbers.
 
SORRY, From b to a, I accidentally replaced the limits with each other in my reply.
Then the ##-## comes from the fact that $$\int_{a}^{b}=-\int_{b}^{a}$$
 
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The work done by a force from a point ##a## to a point ##b## (if the force is conservative) is, by definition $$\int_{a}^{b}\vec{F}\cdot\text{d}\vec{r}$$ (where ##\vec{F}## refers to the applied force. So, the force you need to apply is equal and opposite to the force that is doing the other particle, then the applied force is ##-\vec{F_{e}}## (now ##\vec{F_e}## is the force done by the electric charge) and you have the - you want.

It doesn't make sense to talk about "positive" and "negative" forces, forces are vectors, not numbers.
So you are saying that vectors can't be negative or positive? But I got your answer, Thanks.
 
So you are saying that vectors can't be negative or positive?
Exact, in general, you cannot say that one vector is bigger or smaller than another.
 
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Exact, in general, you cannot say that one vector is bigger or smaller than another.
I'm not saying that, I was just projecting the vectors in a coordinate system
 
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The applied force is opposite to the electrical force, which is ##\vec{F}##. So the applied force equals ##-\vec{F}##. (That doesn't mean it's 'negative' -- that depends on your coordinate system.)
Thanks. I got your answer to my originally asked question. I have another doubt which is does the signs of vectors depend on the coordinate system like the case I stated above.
 
Thanks. I got your answer to my originally asked question. I have another doubt which is does the signs of vectors depend on the coordinate system like the case I stated above.
Vectors don't depend on what coordinate system you choose.
 

Doc Al

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Thanks. I got your answer to my originally asked question. I have another doubt which is does the signs of vectors depend on the coordinate system like the case I stated above.
It only makes sense to say a vector's component along some axis is negative, not the vector itself. And sure, the sign of a component depends upon the coordinate system used.
 
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It only makes sense to say a vector's component along some axis is negative, not the vector itself. And sure, the sign of a component depends upon the coordinate system used.
Got it. That solved my problem.
 

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