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Given this exercise without values into an Italian physics textbook:

Calculates the work required to bring a positive charge from infinity to a finite distance from a negative charge.

The result is a negative work? Why?

Being $W=-\Delta \mathcal U=\mathcal U_i-\mathcal U_f=0-\mathcal U_f$ because $\mathcal U_i(\infty)=0$. The charge are are opposed in sign and I have

$$W=-\mathcal U_f=-k_e\frac{+q\cdot (-Q)}{r}>0$$

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2 Answers 2

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Being $W=-\Delta \mathcal U=\mathcal U_i-\mathcal U_f=0-\mathcal U_f$ because $\mathcal U_i(\infty)=0$. The charge are are opposed in sign and I have $W>0$.

The work required is not $W=-\Delta \mathcal U$ it is $W=\Delta \mathcal U$. If your formula were correct then it would require positive work to lower an object from a shelf down to the floor.

You need to think about the physics and not just blindly plug in formulas. Here, you have a formula for the work and the potential energy but don’t stop to consider if it is the right work. It is best to start with a physical analysis: the force is attractive so the work required is negative. Then you would recognize the sign error immediately.

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    $\begingroup$ When you lower an object from a shelf height $h$ to the floor you do negative work work of $-mgh$ since you exert an upward force of $mg$ opposite to the downward displacement of the object. $\endgroup$
    – Bob D
    Dec 22, 2019 at 22:15
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    $\begingroup$ For my humble opinion the textbook is not a good book. Hence the request $W$ into this book is equal to $\Delta \mathcal U$ or $-\Delta \mathcal U$? And why? $\endgroup$
    – Sebastiano
    Dec 22, 2019 at 22:35
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If you bring a positive and a negative charge together, then force and displacement have opposite direction. So the work is negative.

This may become clearer when you look at the opposite process where the charges are separated. Then force and displacement point in the same direction and the work is positive.

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  • $\begingroup$ Force is attractive because the charges are opposite in sign and I am moving in the direction of the force. $\endgroup$
    – Sebastiano
    Dec 22, 2019 at 22:00
  • $\begingroup$ You must apply a force in the opposite direction. Otherwise the charges were in free fall and you would not end up in a state with a fixed distance. $\endgroup$
    – aventurin
    Dec 22, 2019 at 22:02

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