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## Physics library

### Course: Physics library>Unit 11

Lesson 3: Electric potential energy, electric potential, and voltage

# Electric potential energy (part 2-- involves calculus)

Electric potential energy difference in a varying field. Created by Sal Khan.

## Want to join the conversation?

• why is it a 'minus K' while integrating ?
• because he considered the outward direction to be positive, so the force needed to move it is negative ("radially inward")
• At instead of making the decision to use calculus could he not have used colombs law with r = 10 and then minused it from r=5 to get the energy of being 5m away from q1?
• No. That calculation would give you the difference between the force exerted on the test charge at a distance of 5 meters away and the force exerted on the test charge at a distance of 10 meters away. We want to know how much work it would take to move the test charge from a distance of 10 meters away to a distance of 5 meters away. Work = Force * Distance. We know the distance is 5 meters, because we want to move the test charge 5 meters toward the point charge. However, the force will constantly change as we do this, because the force is inversely proportional to the square of the distance. Therefore, we need to use calculus to account for the continuous change in force as we move the test charge toward the point charge.
• I don't understand what Negative voltage mean... Can anyone explain? This has confused me for a long time
• in simpler words, Voltage(potential) when calculated, is considered to be in the negative direction of Force applied. Thus it is taken negative
• in a medium, what is permittivity and permeability?
• Permittivity has to do with how a material reacts to an electric field and permeability has to do with how a material reacts to a magnetic field.
• A general question about electrical fields: I understand that the magnitude of the field at any given radius is the same, and as the radius increases the magnitude decreases, but does the magnitude ever reach 0?

I would assume obviously not and that it would just become so small that it's negligible, but that implies that EVERYTHING is exerting an electrical field on EVERYTHING in the entire universe... which is mind boggling.
• theoretically field spreads over infinity but as the distance reaches astronomical units it is practically zero
• Why is the work not negative?? At , Sal mentions that the work done to push the positive test charge towards +Q1 should be negative...so after doing the calculus, why don't we get a negative in our answer?? Why does the total work become a positive value of kQq/10?
• The work done to push a positive toward positive is positive. Work is positive when the force is in the same direction as the displacement.

In the problem, he has defined outward as the negative direction. Therefore a force directed inward is a negative force, AND a displacement directed inward is negative. The negative force * the negative displacement will, after the multiplication is done, give you positive work
• Aren't we suppose to swap the bounds since the upper bound, 5 is smaller than the lower bound, 10?
(1 vote)
• No,since we are moving a particle from 10 meters to 5 meters. Lower bound is initial,upper bound is final.
• We know that V= Cq/ r . If r=0 then the value of V reaches infinity. Does that mean we can never make the distance between two particles of positive charges zero?
• Yes. Then they would be in exactly the same place, and that's not possible, right?
(1 vote)
• I think your integral should give you (r^(-1))/(-1), shouldn't it? I could be wrong but I'm almost positive.
(1 vote)
• Yeah that correct. However that is the same as -1* r^(-1), which he got. He then proceeded to move the -1 to the outside of the brackets, and used it to cancel out the other negative. He was a little unclear on that in the video, good question.