IIT JEE , AIEEE Forum - Physics , Electricity

mohile

when one electron is taken towards the other electron,then the electrostatic energy of the system decreases,increases,remains unchanged or becomes zero.

vibhav1991

we know;
F=KQq/square(r)
both of e are of same charge i.e. negative.
so,
q.q>0 (repulsive force)
also distance of separation decreases
therefore:
F>0
now, W=F.Displacement
so
W increases which is stored in the form of energy.
correct me if i am wrong

praveenjoshi007

as u=kqq/r

hence factor "r" will determine this question further as "qq" remains same but the distance is shortning out;

hence energy will increse

rishipratimm

the electrostatic field is a conservative field.
and in a conservative field,the net potential energy of a stable state is always the least.(in this case,electrostatic energy)
now,when two electrons are brought closer to one another,there will naturally be a repulsion and the electrons will try to move away from each other,thus indicating that bringing them closer to one another increases the instability of the system.now,as the stabe state has the least energy,thus when we bring,there is an increase in the electrostatic energy which causes the system to become unstable.

mehrotra.pulkit

The electrical potential energy of a system of charges is equal to the work that must be done to assemble the system.

Let us assemble the system one by one.

1)Bring the charge - q₁from infinity to the specified point.

2)No work done till now as no force is acting on the charge.

3)Now bring the second particle -q₂from infinty and try to push it towards the other charge.

4)As there is repulsion between the two charges we have to do POSITIVE work on the system.

5)This posotive work stores POSITIVE POTENTIAL ENERGY in the system.

6)This energy is given by :-q₂V = [ (-q₂)(-q₁)/4 $\pi$ $\epsilon$ r ] =q₁q₂/4 $\pi\epsilon$ $\pi\epsilon$ r

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