Capacitors, equipotentials, and electric fields

[palaphys]

Homework Statement

refer to the image below. assume that there is a battery connected at P, such that the positive terminal is connected to the wire through P. explain:
a) Is the potential at A, B and C and the branch of this wire at the same potential? Why, or why not?
b) is there an electric field in the wire at this region?

Relevant Equations

$\mathbf{E} = - \nabla V$

I am very confused, all i have thought about right now is that, in that isolated branch connecting A, B and C, the net charge would be conserved. Not sure how to proceed. I feel that

, if there is a charge $-Q$at A, and a charge $-Q_1$ at B, then the charge at C MUST be $Q+Q_1$ maybe somehow, due to these charges, there would be an electric field in the wire. But my mind tells me that there should be no current hence no electric field, due to electrostatic equilibrium. (thereby constant potential?!?) Not sure which is right, and would appreciate any help.

 

[kuruman]

Assuming that all capacitors are uncharged before connecting them to the battery, once they are connected, there will be a transient current until the capacitors are fully charged. When equilibrium is reached, current will stop flowing and the wires connecting the capacitors will be equipotentials.

The isolated "T" shaped piece that you mentioned will be at a higher potential than the wire connecting N to C₃ and C₂ but lower potential than the wire connecting P to C₁. This is accomplished by the battery pulling a certain amount of electrons from the the wire connecting P to C₁ and adding the same amount of electrons to the wire connecting N to C₃ and C₂. As a result the electrons in the "T"-shaped piece are redistributed internally. The entire assembly from N to P is electrically neutral.

 

[palaphys]

Assuming that all capacitors are uncharged before connecting them to the battery, once they are connected, there will be a transient current until the capacitors are fully charged. When equilibrium is reached, current will stop flowing and the wires connecting the capacitors will be equipotentials.

Alright, I understood the equilibrium part, so the electric field in that region is gonna be zero.. is that right?

The isolated "T" shaped piece that you mentioned will be at a higher potential than the wire connecting N to C₃ and C₂ but lower potential than the wire connecting P to C₁. This is accomplished by the battery pulling a certain amount of electrons from the the wire connecting P to C₁ and adding the same amount of electrons to the wire connecting N to C₃ and C₂. As a result the electrons in the "T"-shaped piece are redistributed internally. The entire assembly from N to P is electrically neutral.

alright so that means that the parts connected to N i.e the parts of the capacitor that I have NOT marked, are going to be positively charged. that means that the parts I HAVE marked, i.e B and C are both going to have a positive charge. Hope this is correct.

Also, I do not know much about current electricity, but I am curious to know- is it always the case that in electrostatic equilibrium, or in any other case, the net charge is zero? (I mean sum of the charges when I say net charge)

 

[anuttarasammyak]

Charge conservation of ABC "T" part
-Q_1+Q_2+Q_3=0
Voltage of ABC "T" part measured from N
\frac{Q_2}{C_2}=\frac{Q_3}{C_3}
PN voltage
\frac{Q_1}{C_1}+\frac{Q_2}{C_2}=V

Three equations for three parameters. Isn't it enough ?

 

[palaphys]

View attachment 360218
Charge conservation of ABC "T" part
-Q_1+Q_2+Q_3=0
Voltage of ABC "T" part measured from N
\frac{Q_2}{C_2}=\frac{Q_3}{C_3}
PN voltage
\frac{Q_1}{C_1}+\frac{Q_2}{C_2}=V

Three equations for three parameters. Isn't it enough ?

Yes, solving the questions isn't an issue for me, its more about the concept involved.
I wanted to enquire about the electrostatic equilibrium involved as mentioned in my previous reply

 

[Steve4Physics]

Alright, I understood the equilibrium part, so the electric field in that region is gonna be zero.. is that right?
is it always the case that in electrostatic equilibrium, or in any other case, the net charge is zero? (I mean sum of the charges when I say net charge)

The term electrostatic equilibrium is typically applied to a conductor where there is no electric field inside the conducting material so there is no current. (If the electric field were non-zero, that would cause a current.)

A conductor's net charge does not have to be zero for electrostatic equilibrium.

When electrostatic equilibrium exists, the mobile internal charge-carriers (electrons for metals) have moved around and reached their minimum energy configuration; this corresponds to a constant potential everywhere in the material (including at the surface). Since the potential is constant everywhere, the electric field is zero everywhere ($\mathbf E = - \nabla V$).

A few additional noteworthy points about electrostatic equilibrium...

If there is a net charge, this will all be on the surface.

If there is zero net charge, there can still be surface charges. For example if you place a neutral conducting sphere in a uniform electric field, there will be induced positive surface charges on one side and induced negative surface charges on the other side (but the field inside the conductor will be zero).

Inside the conductor, the charge-distribution may not be uniform – the surface and internal charges will have cleverly redistributed themselves to make the potential constant everywhere. The zero internal electric field is the vector-sum of:
- any external electric field;
- the field from any surface charges;
- the field from internal charges (fixed and mobile).

Any external field lines will be normal to the conductor’s surface (so they have no tangential component at the surface) and start or end on a surface charge.

Edit - minor changes.