Why Is Capacitance Considered Constant in a Conductor?

[Higgsono]

The capacitance C of a conductor is given to be a constant relationship between charge Q and potential V of the conductor given by Q = CV.
But how can C be a constant? Because the potential of the conductor will not be a linear relationship of the charge that I add. THe more charge there is on the conductor already, the more work is needed to add additional charge. Hence the potential I add by bringing new charges to the conductor must depend on the charge already there.

What is it that I don't understand?

 

[Dale]

Because the potential of the conductor will not be a linear relationship of the charge that I add.

The potential is linear with the charge that is on the conductor, not the charge that is added. I.e. $V$ is proportional to $Q$ not $\Delta Q$

 

[Higgsono]

The potential is linear with the charge that is on the conductor, not the charge that is added. I.e. $V$ is proportional to $Q$ not $\Delta Q$

huh? Q and V are not constants. I must be able to double the charge and the relation should be the same right?

 

[Dale]

huh? Q and V are not constants. I must be able to double the charge and the relation should be the same right?

Yes, the proportionality between $Q$ and $V$ is constant, but what you are describing in your text is the relationship between $\Delta Q$ and $\Delta W$. $\Delta Q\ne Q$ and $\Delta W \ne V$

 

[VicomteDeLaFere]

Remember that C is a physical quantity. Remember back to the simple capacitance days and how two plates make a capacitor. The capacitance depends on dimensions and material properties. It does not depend on electrical properties. It is a constant as long as the materials and the physical dimensions don't change.

As for the charge and voltage, capacitance is the ratio of the two, C=Q/V; therefore Q and V don't have to be linear, they can follow any line as long as the ratio between them remains constant.

If you want to look at the different lines that Q and V follow, look at charging and discharging a capacitor at constant current versus constant voltage/resistance.

 

[Dale]

Let’s try it more quantitatively

Because the potential of the conductor will not be a linear relationship of the charge that I add.

That is correct. $V \propto Q$ not $V \propto dQ$

THe more charge there is on the conductor already, the more work is needed to add additional charge.

Yes, $dW/dQ=f(Q)$ where $df/dQ>0$

Specifically
$dW/dt=P=IV=V \; dQ/dt$
$dW/dQ = V = Q/C =f(Q)$
So $df/dQ =1/C>0$

Hence the potential I add by bringing new charges to the conductor must depend on the charge already there.

No, it does not follow. Instead $dV/dQ = 1/C$