Potential energy of a System of charges

[Physicslearner500039]

Homework Statement

Two electrons are fixed 2.0 cm apart. Another electron is shot from infinity and stops midway
between the two. What is its initial speed?

Relevant Equations

NA

This is my attempt the system

The 1 is the initial configuration where the 3 electron is at infinity.

The 2 is the final configuration where the 3 electron is midway.U1 is the potential energy between e1 and e2
U1 = (q1*q2)/(4*π*ε0 * (0.02)^2); // q1, q2 charge of electrons
K1 = 0.5*me*V^2; // me mass of electron, V initial velocity to find out.

In configuration 2:

Potential energy = U12 + U13 + U23;

U12 = U1;
U13 = U23 = (q1*q2)/(4*π*ε0 * (0.01)^2);

U1 + K1 = U12 + U13 + U23;
K1 = 2*e2/(4*π*ε0 * (0.01)^2);
I then find the initial velocity, Is my understanding correct? Please advise.

 

[BvU]

U1 = (q1*q2)/(4*π*ε0 * (0.02)^2); // q1, q2 charge of electrons

[edit]:
Try again. From dimensions you can see this isn't right My bad --- $U$ is an energy, not a potential. The notation confused me.

I am almost confused by a renaming action like $U_1 = U_{12}$ -- Why introduce this ? Next line, you write $U_1 ... = U_{12} ...$ again, as if you want to conceal that they are one and the same.

Is my understanding correct?

Why the doubt ?

PS do you think this is a realistic exercise ? Why / c.q. Why not ?

 

[ehild]

Homework Statement:: Two electrons are fixed 2.0 cm apart. Another electron is shot from infinity and stops midway
between the two. What is its initial speed?U1 is the potential energy between e1 and e25

In configuration 2:

Potential energy = U12 + U13 + U23;

The electrons 1 and 2 are fixed. only the potential energy of the third one counts., the energy of the third electron is conserved.

 

[Neelima]

Homework Statement:: Two electrons are fixed 2.0 cm apart. Another electron is shot from infinity and stops midway
between the two. What is its initial speed?
Relevant Equations:: NA

View attachment 261125

This is my attempt the system

View attachment 261127
The 1 is the initial configuration where the 3 electron is at infinity.

The 2 is the final configuration where the 3 electron is midway.U1 is the potential energy between e1 and e2
U1 = (q1*q2)/(4*π*ε0 * (0.02)^2); // q1, q2 charge of electrons
K1 = 0.5*me*V^2; // me mass of electron, V initial velocity to find out.

In configuration 2:

Potential energy = U12 + U13 + U23;

U12 = U1;
U13 = U23 = (q1*q2)/(4*π*ε0 * (0.01)^2);

U1 + K1 = U12 + U13 + U23;
K1 = 2*e2/(4*π*ε0 * (0.01)^2);
I then find the initial velocity, Is my understanding correct? Please advise.

Method is correct. But you should not square the position. It is like U= ( K x q1 x q2) / r
I think do

 

[rude man]

Equate the potential energy of the third charge at the midpoint to the kineic energy at infinity. Looks like you're trying to do that. Also note what neelima said about r. I don't understand why you have all those $U_{ij}$ terms. There is only one potential energy, and one kinetic energy.

 

[rude man]

(For extra credit:) The third electron sits exactly between the two stationary electrons. The third electron is then displaced a small amount directly towards one of the stationary electrons. What is the frequency of oscillation?