# how could Kinetic energy increase if no energy was there in the system?

i was reading Physics-Halliday, Resnick, Krane(fifth edition) there on page 638 it is written:

if the external agent does positive work in assembling the charge from the infinite separation (opposing a repulsive force in the process), the total potential energy calculated will be positive. The external agent has in effect stored energy in the system of charges. If the charges are released from their position, they will tend to fly apart, the potential energy will decrease and the kinetic energy will increase

so i assumed the following situation:

(the above situation is assumed in a vacuum space with no other force there)

suppose in the region A has equidistant parallel electric lines of electric field which is due to a positively charged body somewhere in the space but not at infinity from region A and constituting the system and the other being a small positively and uniformly distributed charged body(B) which is slowly moving with a velocity but with no acceleration but hasn't entered yet in the region A

B is moving towards the region A from infinity(initial position) with zero acceleration hence it has done no work yet now when it would just enters the region A perpendicularly.

So when it will move just inside the field it would have moved a distance $dx$(final position) perpendicular to the directio of the electric field so the work done again will be zero.

so from moving from infinity to this final position the total work is zero hence $$U_i-U_f=0\implies U_i=U_f=0$$

*1)furthermore if no work is done on the system then no potential energy stored in the system then when the body B will set free then there will not be any energy (potential) which sould be converted into kinetic energy hence in this case body B should not increase its KE.

*2)But you see, although body B has traveled a distance $dx$ but it has traveled a distance so the body other than B would exert a force on the body B which is in the direction of $\bar{E}$ which would made it to accelerate it in the direction of $E$ and hence it would increase its kinetic energy.

now you see that firstly(*1) i said that there is no potential energy would generate in the system because the external agent has done no work on the system and hence there will not be any change in KE but in the other hand(*2) i said that the body B acquire the KE.

In short how could KE increase if no energy was there in the system?

but we all know that energy can't be generated so there is something which i am taking wrong.

• Why are you assuming that b is not accelerating? It will experience the electric field regardless of how far away it is, The force lines you re showing don't reflect what the actual electric field of those charges look like. They are the field between the charges but the field extends in all directions. – Peter R May 5 '16 at 13:43
• As drawn, the electric field is 'peculiar' for a "positively charged body somewhere in the space". There is clearly non-zero curl at the horizontal boundary of region A and, thus, the electric field you have drawn is not conservative. That is, we can't define a potential difference between two points, one inside region A and one outside since the work done is path dependent. – Alfred Centauri May 5 '16 at 14:01
• I might say that the field that you draw is impossible to produce using only fixed charges to produce it. You can't have a region of space where there is no field. I think I agree with @AlfredCentauri, and his analysis sounds more general than mine, but I'm not sure I fully understand his short comment. – garyp May 5 '16 at 14:17
• Please take a look at all the comments. The situation as you present it is not clear. If you are not satisfied with the answers (and the answers in the comments) please edit the question to make it more clear. – garyp May 5 '16 at 15:18

Electric potential energy is stored in your system of a large mass charged body $A$ and a small mass charged body $B$.

When an external force is applied to body $B$ that external force will do work on the system (body $A$ and body $B$) and change the electric potential energy of the system and possibly change the kinetic energy of body $B$.

However it is often the case that the external force is made equal and opposite to the force on body $B$ due to body $A$ which is the force on body $B$ due to the electric field produced by body $A$.
The consequence of that is that for body $B$ there is no net work done on it and its kinetic energy does not change.
In such a case the external force has done work on the body $A$ and body $B$ system and in the process the kinetic energy of body $B$ has not changed but the potential energy of the body $A$ and body $B$ system has changed.

Now release body $B$ whilst it is the electric field produced by body $A$.
Considering only body $B$ as the system you can say that body $B$ has an external force on it due to body $A$ and that external force does work on body $B$ which changes the kinetic energy of body $B$.

If you consider the system comprising body $A$ and body $B$ then there can be no change in the total energy of the system.
What happens is that the potential energy of the system $U$ changes from $U_i$ to $U_f$ and the kinetic energy of body $B$ changes from zero to $K_f$.
$$U_i +0 = U_f + K_f$$

As with a very similar analysis of raising a mass (body $B$) in the Earth's (body $A$) gravitational field I have assumed that the mass of body $B$ is much, mush smaller than that of body $A$ and so the change in kinetic energy of body $A$ is negligible.

Apparently you are not considering the work done by the electric field, and you are confusing yourself with the "energy storage" concept.

After traveling a $\Delta x$ distance into the electric field $\vec{E}$ (avoid using $dx$ because $d$ usually stands for infinitesimal changes), the particle B with charge $q$ will have suffered change in its "horizontal" given by the Work-KE Theorem:

$$W_{net} = \Delta KE \longrightarrow E \cdot q \cdot \Delta x= \frac{m}{2} \Delta (v^2)$$

"Vertical" velocity remains unchanged because the force ehxerted by the electrical field is always perpendicular to the initial velocity. Thus the final KE is

$$KE_{final} = E \cdot q \cdot \Delta x + \frac{mv^2}{2}$$

The energy was stored in the electric field, i.e. stored in potential energy on the particles of the charged body. The approximation made is that $\Delta KE$ of the body B produces no apparent change on the energy of the charged body that produces A. This could be done by holding it still with some apparatus, or by using an extremely massive body (so that changes in B wouldn't produce notable changes on the charged body).

• I may be wrong, but I think you missed the point that in the OP's set up, he's imagining that it took no work to bring particle B into the system, and hence there was no potential energy associated with it. – garyp May 5 '16 at 14:12
• Possibly, but on 2* he says that there will be no change on KE after B ventures into A. This false, as of my answer. – big-lion May 5 '16 at 14:28
• Yes, the way he phrased it is not true. I went out on a limb again and assumed that he meant that there will be no change at the instant in enters into area A. All things considered, the question is not clearly stated. – garyp May 5 '16 at 15:15

Perhaps the confusion comes from the often belief that if a force field is only a function of position then the force is conservative, and that non-conservative forces are those that include friction dependence on speed, and other variations. But this is incorrect, as alfred centauri commented this field is not conservative. But this is not just because there is an unphysical discontinuity at the border. Even if you smooth the field to keep it parallel but making it go to zero smoothly (for instance, using $E_x=1+\cos y$ on the transition, you will never find a potential function, the total work made by the force will be a function of the path in that transition layer.