# Increase of potential when charging a capacitor

I have seen previous posts on why the potential of a fully charged capacitor is equal to that of the external battery. But I was not satisfied by the analogies made that charges will repel each other or that the capacitor will oppose the battery from more charges to be flowed. But how can we prove with the idea of electric fields and forces that the capacitor will oppose the battery from accumulating further charges? The polarity of a capacitor is same as that of the battery. So,both the electric fields of the capacitor and the battery point in the same direction,thus adding up a higher electric field within the capacitor. If the capacitor were to oppose the battery,wouldn't the electric fields have to point in opposite directions so that net electric field becomes 0? A solution addressing this problem will be much appreciable.

In fundamentals of physics book by Resnick and Haliday it has been written that the electric field of the battery drives electron from the positive plate of the capacitor and pushes electrons to the negative plate. At one point when the potential of positive terminal of battery becomes equal to the potential of positive plate of capacitor,there won't be any electric field in the wire to push charges. But in circuits,we know that all the points are in same potential in a resistance less wire,still current flows through the wire. In this case also the two potentials have become equal. So why no flow of current in this case?

• See eng.libretexts.org/Bookshelves/Electrical_Engineering/…’s_Voltage_Law_for_Electrostatics_-_Integral_Form Commented Nov 8, 2023 at 12:06
• Look up kirchhoff’s voltage law for electrostatics Commented Nov 8, 2023 at 12:20
• Just like the algebraic sum of the voltages in a loop is zero, so is the sum of the electric fields zero Commented Nov 8, 2023 at 12:33
• The battery doesn’t add to the electric field of the capacitor Commented Nov 8, 2023 at 12:35

Here is the circuit when current has ceased to flow with some extra annotations.

Let us take unit positive charge around the circuit $$XY(Z)X$$,

$$X\to Y$$ : Work done by external force is $$\displaystyle \int_{\rm X}^{\rm Y}- \vec E_{\rm B}\cdot d\vec \ell = \int_{\rm X}^{\rm Y} +E_{\rm B}\,d \ell=V_{\rm XY}$$ the potential of node $$Y$$ relative to node $$X$$ and note that the external force does a positive amount of work.

$$Y\to X$$ via node $$Z$$ : Work done by external force is $$\displaystyle \int_{\rm Y}^{\rm X}- \vec E_{\rm C}\cdot d\vec \ell = \int_{\rm Y}^{\rm X} -E_{\rm C}\,d \ell=V_{\rm Y(Z)X}$$ the potential of node $$Z$$ relative to node $$Y$$ and here the work done by the external force is negative.

The total work done by the external force in moving the charge round a complete loop is $$V_{\rm XY} + V_{\rm Y(Z)X}=0$$ by energy conservation.

However $$V_{\rm XY}= -V_{\rm YX}$$ and so $$-V_{\rm YX} + V_{\rm Y(Z)X}=0\Rightarrow V_{\rm YX} = V_{\rm Y(Z)X}$$, the potentail difference across the terminals of the cell is equal to the potential difference across the plates of the capacitor.

• Thank you for answering,but while going from $Y$ to $X$ via $Z$,didn't we assume that the electric field through out is $E_C$? But isn't the electric field of a capacitor only contained within the interior of the capacitor? More over,I can't understand why in the $YZX$ loop,the battery's electric field doesn't have any contributions. It will be very helpful if you answer this. Commented Nov 8, 2023 at 13:38
• I have assumed there is no potential difference across the wires as there is no current current passing through them,. That being the case as the electric field is minus the potential gradient there is no electric field there. Commented Nov 8, 2023 at 14:58

So, both the electric fields of the capacitor and the battery point in the same direction, thus adding up a higher electric field within the capacitor.

The electric field in a parallel plate capacitor, neglecting edge effects, is given by

$$E=\frac{V}{d}$$

Where $$V$$ is the voltage across the plates and $$d$$ is the plate separation distance.

Say you have an isolated charged capacitor with voltage $$V$$. Connecting it in parallel to a battery of voltage $$V$$ doesn't change the voltage across the capacitor. Therefore the presence of the battery doesn't change the electric field of a fully charged capacitor, where "fully charged" means the voltage across the capacitor is the battery voltage.

Hope this helps

• Thank you for answering,could you be available for a chat room for answering some of my queries if you don't mind? Commented Nov 8, 2023 at 14:26
• @a_i_r Sure. You set it up Commented Nov 8, 2023 at 14:43
• chat.stackexchange.com/rooms/149579/physics kindly join. Commented Nov 8, 2023 at 14:49
• Extremely sorry,it seems I can't send any message,so I am sharing my problem here. Commented Nov 8, 2023 at 14:56
• We have a simple circuit consisting of a battery with voltage $V$ and resistance $r$. Now all the points along the wire till the resistance are at the same potentials and this potential changes only across the resistance. But we know current obeying ohm's law flows through this circuit. Hence current also flows through those similar potential points. Commented Nov 8, 2023 at 14:57