Physics Help Forum Positive work where negative work should be

 Electricity and Magnetism Electricity and Magnetism Physics Help Forum

 Mar 24th 2013, 05:42 AM #1 Member   Join Date: Jul 2012 Posts: 67 Positive work where negative work should be Two positive charges of q=3*10^-6 C are a distance 1 m apart. What will be the work done by the electrical force in bringing one of the charges to a distance of 0.5 m from the other charge (which is fixed at its position through the whole process)? Well from W=-ΔU I got w=-0.081 J which makes sense because the electric force acts in contrast to the path of q as it moves closer to the other q - and so negative work is done. But when I tried to calculate the work directly from W1m->0.5m = ∫Fel*dr when bottom limit=1m and upper limit=0.5m , I got the same result but positive - since the angle between Fel and dr is 180 deg, you multiply this minus 1 with the minus 1 that comes out when you calculate the integral of 1/r^2 (which is -1/r, and if you put in the integration limits you'll get -1 as I noted) - kq1q2 is a positive number. So why is that? Isn't dr pointing to the direction of heading? It seems that when I do the transition from dr the vector to dr the scalar I'm suppose to write: -dr and not dr but I can't understand why and what does this minus sign represent...
 Mar 25th 2013, 07:37 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,321 The problem is the extra negative sign you introduced in the integral, based on the direction of force being opposite the direction of movement. No need to do that, as the limits on the definite integral take care of that for you. Remember that positive force is measured in the positive r direction, so no need to add the extra negative sign. By adding in that negative sign you are effectively calculated the work done on the electric field by the particle, not the work done on the particle by the electric field. Last edited by ChipB; Mar 25th 2013 at 10:00 AM.
Mar 25th 2013, 09:12 AM   #3
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 Originally Posted by ChipB The problem is the extra negative sign you introduced in the integral, based on the direction of force being opposite the direction of movement. No need to do that, as the limits on the definite integral take care of that for you. Remember that positive force is measured in the positive r direction, so no need to add the extra negative sign. By adding in that negative sign you are effectively calculated the work done on the electric filed by the particle, not the work done on the particle by the electric field.
What do you mean by that the limits take care of that for me? From what I know - dr is the very small displacement vector that points in the direction of heading, that means from 1m towards 0.5m (towards the stationary charge). F is the electrical force on the moving charge and it acts outwards (repulsion) so the angle between F and dr at all time is 180 deg. These facts are true even if I wasn't to do integration at all, aren't they?

 Mar 25th 2013, 10:33 AM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,321 We need to be careful about calculting the work done by teh aprticle versis the work done by the electric field. Starting with work done by the particle - it's defined as: where the vector F points in the direction of the force that the particle is applying to the electric field. Let's assume that the particle is moving from left to right (positive x direction) and the force being applied by the particle to the electric filed is in the positive direction. The starting position of the particle is at -1 m and its final position is at -0.5m. The integral becomes: Now let's take the point of view of work by the electric field. For a particle approaching from the left moving from -1m to -0.5m in a direction opposite to the force applied by the field you get Alternatively, for a particle approaching from the right: Note that the sign of the force is defined as applied by the object whose work you're trying to calculate.
Mar 25th 2013, 11:55 AM   #5
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 Originally Posted by ChipB We need to be careful about calculting the work done by teh aprticle versis the work done by the electric field. Starting with work done by the particle - it's defined as: where the vector F points in the direction of the force that the particle is applying to the electric field. Let's assume that the particle is moving from left to right (positive x direction) and the force being applied by the particle to the electric filed is in the positive direction. The starting position of the particle is at -1 m and its final position is at -0.5m. The integral becomes: Now let's take the point of view of work by the electric field. For a particle approaching from the left moving from -1m to -0.5m in a direction opposite to the force applied by the field you get Alternatively, for a particle approaching from the right: Note that the sign of the force is defined as applied by the object whose work you're trying to calculate.
I'm interested in your last approach - the work by the electric field (or of the electrical force on that moving particle) - are you saying that the angle between the electrical force on the moving particle and dr is 0?? Why aren't you treating the situation like when calculating the work friction does: you look at the friction force acting on a body, it seems to me that you don't look at the electrical force that acts on the moving charge...

 Mar 25th 2013, 12:11 PM #6 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,321 Let's consider the work done by friction. If you push on a box across the floor in the positive x direction with force F to overcome friction forces, and the box moves distance D in the positive x direction then the work you do is +FD (absoulte value of force times absolute value of distance times cosine 0). But from the point of view of the floor the force it applies to the box is in the negative x direction, so the work done by the floor on the box is -FD (absolute value of force times absolute value of distance times cosine 180). And yes - the angle between the electric force and dr is zero in the last case is 0, because the force is in the positive direction and dr is measured in the same direction. To see that this works - suppose the particle was moving away from the fixed charge, so the the integral would then be from 0.5m to 1m, and you would get a positive value for work done by the electric field on the moving particle, as expected. So if switch direction of movement you flip the limits of the integration, but you don't throw a negative sign in, unless dr is measured in the opposite direction from the +F direction. Last edited by ChipB; Mar 25th 2013 at 12:28 PM.
Mar 25th 2013, 12:50 PM   #7
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 Originally Posted by ChipB Let's consider the work done by friction. If you push on a box across the floor in the positive x direction with force F to overcome friction forces, and the box moves distance D in the positive x direction then the work you do is +FD (absoulte value of force times absolute value of distance times cosine 0). But from the point of view of the floor the force it applies to the box is in the negative x direction, so the work done by the floor on the box is -FD (absolute value of force times absolute value of distance times cosine 180). And yes - the angle between the electric force and dr is zero in the last case is 0, because the force is in the positive direction and dr is measured in the same direction. To see that this works - suppose the particle was moving away from the fixed charge, so the the integral would then be from 0.5m to 1m, and you would get a positive value for work done by the electric field on the moving particle, as expected. So if switch direction of movement you flip the limits of the integration, but you don't throw a negative sign in, unless dr is measured in the opposite direction from the +F direction.
Bud isn't dr pointing towrds the movement path? Because if it is - then it's in the opposite direction of the force... Is there another convention where dr points regardless to the direction of movement?

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