What is the equivalent capacitance between points A & B?

[Kronos]

Homework Statement

Find the equivalent capacitance between points A & B

Homework Equations

Ceq₁ = C + C
1/Ceq₂ = 1/2C + 1/2C + ...

The Attempt at a Solution

[/B]
It is my understanding that each capacitor top and bottom are connected in parallel, so their equivalent is 2C.

Now, they are all connected in series, so their equivalent would be:
1/Ceq₂ = 1/2C + 1/2C + 1/2C + 1/2C

Thus,
Ceq₂ = C/2

Is my reasoning acceptable?
Thank you.

 

[haruspex]

each capacitor top and bottom are connected in parallel

If the effective capacitance sought were between the left hand side and the right hand side, that would work. But it isn't.

Is there more context here, like an applied varying potential?
If not, think about potentials. What might the potential difference be between the right hand side of the top left capacitor and the right hand side of the bottom left capacitor?

 

[Kronos]

The potential difference would just be zero since the capacitors are at steady state. Thus, there is no current flowing through the wire w/ resistor, and by Ohm's law:
V = I*R
V = (0) * R
V = 0

 

[haruspex]

The potential difference would just be zero since the capacitors are at steady state. Thus, there is no current flowing through the wire w/ resistor, and by Ohm's law:
V = I*R
V = (0) * R
V = 0

Right. Does that help you simplify the circuit?

 

[Kronos]

Right. Does that help you simplify the circuit?

Yes. That means that the capacitor on the top left and the capacitor on the bottom left are connected in series, correct? Since the same charge will flow through them.

 

[haruspex]

Yes. That means that the capacitor on the top left and the capacitor on the bottom left are connected in series, correct? Since the same charge will flow through them.

That's how I read it.