# Capacitance of 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?

Mentor
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?

Mentor
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.

Mentor
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##