# The electric field between two plates of a capacitor

Suppose you a have a parallel plate capacitor connected to a battery of voltage $$V$$, and charge is starting to accumulate on the side of the capacitor connected to the positive end of the battery.

Just a very short time after the process starts, what's the potential difference between the two sides of the capacitor ?

I'm getting two different answers from two different sources, and I'm being confused:

• If the potential difference between the two plates is $$\frac{Q}{C}$$, where $$C$$ is the capacitance, and $$Q$$ is the present charge on the capacitor, then at the very beginning, $$Q = 0$$ so there's no potential differnce. Then why would the charge flow from one plate to the other plate without any potential differnce ? From this I get the impression that there should be a electric field at first for the charges to even start moving.

• Shouldn't the potential between the two plates be actually $$V$$ since the wires and the plates are conductor and the potential is same across the conductor; so the potential difference between the two plates should be the potential difference between the two sides of the battery which is $$V$$ ?

Just a very short time after the process starts, what's the potential difference between the two sides of the capacitor ?

At first V=0.

Then why would the charge flow from one plate to the other plate without any potential differnce ?

You are mentally treating a capacitor as though it were a resistor with this statement. For a resistor $$V=IR$$ so indeed you do need to have a voltage across a resistor to have charge flow (current). However, that is not how a capacitor behaves. For a capacitor $$I=C \frac{d}{dt}V$$, so current flows through a capacitor not based on how much voltage is across the capacitor, but based on how quickly that voltage is changing. The voltage is 0, but it is changing, so there is a current.

It is important to not make the mental mistake of treating other components as though they were resistors.

the potential difference between the two plates should be the potential difference between the two sides of the battery which is V ?

This would be true if the battery were an ideal voltage source. The current would therefore be infinite since an instantaneous change in voltage across a capacitor requires an infinite current. Since no real battery can source an infinite current, you would have the maximum current determined by the battery’s internal resistance which is sometimes called the short circuit current. That would increase the voltage at the maximum rate possible, but not an instantaneous jump.

I see Dale has given you a good qualitative answer so I will give a more mathematical one.

The circuit you are considering is a $$RC$$-circuit which you can solve for the charge $$Q$$. The differential equation reads $$\frac{dQ}{dt} = -\frac{Q}{RC}$$ where $$R$$ is the resistance of the wires. The solution of this equation is $$Q_{source} = Q_0 e^{\frac{-t}{RC}}$$ which gives the charge at all times at the source. You can of course rewrite this for the charge in the capacitor because charge has to be conserved : $$Q_{capacitor} = Q_0(1- e^{\frac{-t}{RC}}).$$ So the potential difference between the two plates of the capacitors at any time $$t$$ is $$V_{capacitor} = \frac{Q_0}{C}(1- e^{\frac{-t}{RC}}).$$ You can see that at time zero there is no charge or potential difference in the capacitor and if you wait long enough, then there will be a potential difference of $$V$$.