Electric Force to Electric Potential Energy - Why not both linear?

Question

Why is electric force from an electric field on a test charge increase exponentially with reduction in the distance between the test charge and the source, but the increase in electric potential energy linear with decrease in difference? (Assuming a negatively charged source)
Is it because the force increase is 'split' between the source and the point charge? Does the source charge also get an increase in potential energy? I'm assuming a simple field with a static source charge.

Why does the force go up much faster than the energy of the test charge as it approaches the source?

I'm looking for a conceptual explanation, I know how to manipulate the equations, but I want to understand it beyond the math.

Thanks

Answer 1

Difficult questions! And very hard to explain properly without calculus. Without maths, it makes my answer quite long. But here goes...

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Force is inversely proportional to distance squared. That's because the effects of a charge spread out in 3D. If you triple the distance, the effect is spread out over 3² = 9 times the area. So the force is 9 times weaker if the distance is tripled.

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"electric potential energy [is] linear with decrease in difference"



Not quite. Potential energy is actually inversely proportional to distance. If distance changes by a factor f, potential energy changes by a factor 1/f.



E.g. if distance triples, potential energy becomes 1/3 of its initial value.

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The signs (+ or -) do not affect the inverse proportionality. So there is no need to consider a 'negatively charged source'.



For the explanation below, I'll take it that we are talking about 2 positive charges.

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"Does the source charge also get an increase in potential energy?"



That's not the correct way to think about it.



It is best to think of the potential energy as being stored in the electric field. Or stored by the whole system. Or stored by the *pair* of charges. The potential energy doesn't belong to either charge alone.



But for simple calculations we often get away with treating the potential energy as 'owned' by only one charge. That's because we often have a fixed charge, so changes in a system's potential energy results in a kinetic energy change for only the charge able to move.

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The big difference (apart from their vector and scalar natures) between electric force and electric potential energy is this:



a) Force is a value at a given point.



b) Potential energy is actually a difference in energies between 2 points. It is the work done moving between 2 points. We can choose any convenient start point to call zero energy. Then potential energy at some point P is the work done moving from the reference point to P.



The reference point (zero potential energy) is usually taken as 'infinity'.



Suppose we have a fixed charge A.



The potential energy at point P, a distance d from A, is the work done moving charge B from the reference point to P.



To move charge B from the reference point (infinity, where potential energy is zero) to a point some distance d from a charge B, requires energy (work) to overcome the repulsion.



But *the average force* while moving A between infinity and P is much less than the force at P.



It is the fact the *average force* to get to P is much less than the force at P, that gives rise to to the difference that bothers you.



The force 2m from A may be 100N.

Then force 1m from A is 400N, four times bigger (inverse square law).



KEY POINT COMING UP!

But the ***average**** force while moving from infinity to 1m is only twice the average force while moving from infinity to 2m.



As P gets closer to A:

- the force at P increase in proportion to 1/d²

- the average force between infinity and P increases in proportion to 1/d, which means potential energy is proportioanl to 1/d.



But you need calculus to show this properly.

Answer 2

First off, the force doesn't decrease exponentially but quadratically. Also, the static source charge can also be the moveable point charge and vice versa, if you just change your reference. We usually use a static charge to simplify things, but of course it also feels the electric field of the point charge.



To give a conceptual explanation is difficult, but I'll do my best.

Whenever you see some (usually) conserved quantity radiated outward from a point with an 1/r^2 dependence, it is best to think of the inverse-square law [1].

It basically says that because this radial outwards behaviour, the quantity must be "smeared out" over an increasingly larger sphere around the source. And since the surface of a sphere goes with r^2, the quantity decreases with this r^2.

The case with electric potential energy (same as with gravitational potential energy) is a bit different. Potential energy is the "stored energy" of an object, due to several factors (like its relative position) and as such is not a vector. In this case it is the energy difference between the energy of the object in its current position and the energy at a reference position (which usually is infinitely far away).

It is easier to understand energy if you relate it to work: how much energy do I need to push a stone 1m. Now how much energy do I need to push it 2 meter? Double the energy of course: once to push it the first 1m, once to push it the second 1m. This I hope feels rather intuitive. This doesn't really involve potential energies, but I could compress a spring (giving it potential energy) and use it to push a small rock 5 cm. I could compress the spring again, and again push the small rock 5 cm: twice the distance needs twice the energy, 1/r behaviour.

Answer 3

sorry, i think the reduction in distance reduces the electric potential energy but increases kinetic energy, which was what I meant to say.