How can kinetic energy decrease and increase at the same time?

Question

To move a positive test charge $q$ towards another positive charge $Q$ one has to apply external force against the repulsive force. To move (q) against nature, the external force should be slightly greater than repulsive force, hence charge (q) will have some slight acceleration therefore it's velocity should increase per second very slightly and hence the kinetic energy also. But we are moving the charge (q) against nature so it's potential energy should increase and kinetic energy should decrease by the same amount so that total energy remains conserved....but how can kinetic energy increase and decrease at the same time? since velocity of the charge is also increasing slightly therefore kinetic energy should increase but by energy conservation the kinetic energy should decrease!

NOTE: here we are moving the charge (q) such that the maximum amount of work done on it should be stored as potential energy and not kinetic energy hence here external force should be ALMOST equal to repulsive force hence i have used the term 'slightly' for specifying that the external force is just a little bit greater than repulsive force and therefore the acceleration is also NEARLY zero but not exactly zero.

Answer 1

You are using an external force to move the charge. The whole problem is that in some of your considerations, you say there is an external force, but when you calculate the energy of the system, you ignore the external force.

There are three kinds of energy here - potential energy between the two charges, kinetic energy of the charge you move, and energy used to apply this external force. Energy is lost from whatever is providing the external force. Some of that energy is converted to kinetic energy, and some of that energy is converted to potential energy.

Another way of saying this, the external force does work on the charge it's moving.

Answer 2

From what I know it is impossible for the kinetic energy of a system to increase and decrease at the same time. That is like taking the equation s=ut+1/2(at2) and then saying that time can be negative and positive (but it can't be -ve). Acc to the problem: It is given that charge q is given external force against the resistive force hence the resultant will be against nature (I agree) But the resistive force will be cancelled hence only there will be kinetic energy in the charge(q)... I think ur having problems in deciding the system cause if u only take the charge (q) the kinetic energy is increasing and the potential energy is decreasing but if we talk abt the system of 2 particles being q and Q then as they come closer the resistive force increases and the potential will increase (still that force will be cancelled out as described. But talking abt the single particle the kinetic energy will keep increasing but as it approaches the other particle it may start to decrease if the resultant force is constant. BTW. And if kinetic energy is increasing and decreasing then potential energy will have no meaning cause then kinetic energy can serve as both.

Answer 3

Look at the system of two charges as a whole with an external force acting on the system.
The two charges start from rest and finish up at rest thus, the work done on the system by an external force is equal to the change in the electric potential energy of the system.

A possible scenario is that at the start the external force is greater than the electric force between the two charges and there is an acceleration (positive) of one of the charges and thus an increase in speed and kinetic energy but then towards the end of the motion if the external force is less than the electric force between the two charges and again there is an acceleration (negative) of one of the charges and thus a decrease in speed and kinetic energy with the increase in kinetic energy at the start just equal to the decrease in kinetic energy at the end.

Thus the net work done by the evternal force is equal to the change in electric potential enegy of the system.

Answer 4

I think use of external force in this scenario is just to simplify the question to avoid use of negative sign,since the direction of external force is opposite to the electric field and its magnitude is equal to electric field. $$\vec{F}=-\nabla.U$$ $$\int{dw_{field}}=-\int{dU}$$ $$Work_{field}=-q\int{dv}$$ If a charge moves with certain velocity towards a charges, then field will do negative work on the moving charge till its velocity becomes zero. This increases the potential energy because the force field is conservative. But the use of infinitesimally greater external force is to get rid of negative sign and to impart some non-zero velocity, which is insignificant in comparison to original velocity. This way calculations aren't much affected. $$\int\vec{F}_{ext}.dr=\int({\vec{F}_{field}+dF}).dr$$ $$\int|\vec{F}_{ext}||dr|cos(0)=\int(|{\vec{F}_{field}|drcos(180°)+dF.drcos(180))}$$ The 2nd term becomes almost zero and $$work_{ext}=-work_{field}$$

This tells us that external force is just used to make calculation easier.

Answer 5

To move (q) against nature, the external force should be slightly greater than repulsive force

This is not true. You can have motion against forces and you can have motion without forces. What is true is that if the forces are unbalanced, you will have acceleration.

But we are moving the charge (q) against nature so it's potential energy should increase.

It is moving against a conservative force, so the potential energy increases.

and kinetic energy should decrease by the same amount so that total energy remains conserved

That is only true if there is no external force. Total energy is not conserved when the external force is present.

Without knowing the specifics of the forces, it is not possible to determine if kinetic energy is increasing, decreasing, or constant. All are possible with different values of an external force.