# How do capacitor circuits work?

im not sure if this is right but when capacitors charge does the postive terminal gains electrons from the capactior and in the wires due to attraction of opposite charges and the capaitor looses these electrons. Therefore the capcitor becomes positively charged and the terminal reudces its postive charge. Does this stop when the potential difference between the capacitor plate and postive termina is 0 and therefore the terminal has lost 6v and the capacitor has gained six volts or am i getting charge and voltage mixed up.

im pretty sure this isnt right
Original post by charlie.R.12
im not sure if this is right but when capacitors charge does the positive terminal gains electrons from the capacitor and in the wires due to attraction of opposite charges and the capacitor loses these electrons. Therefore the capacitor becomes positively charged and the terminal reduces its positive charge. Does this stop when the potential difference between the capacitor plate and positive terminal is 0 and therefore the terminal has lost 6v and the capacitor has gained six volts or am i getting charge and voltage mixed up.

im pretty sure this isnt right

Hi, I provide a link (below) to a pdf file that clears half of your doubt. After reading the material, post what is cleared and what is left uncleared.
Let's take one step at a time.
https://dokumen.tips/documents/capacitance-and-dielectrics-mytextbookswheniwantem-capacitance-and-dielectrics.html?page=2

Read section 26.1 mainly, beginning with “Let’s consider a capacitor formed from a pair of parallel plates as shown in Figure 26.2.”
Original post by Eimmanuel
Hi, I provide a link (below) to a pdf file that clears half of your doubt. After reading the material, post what is cleared and what is left uncleared.
Let's take one step at a time.
https://dokumen.tips/documents/capacitance-and-dielectrics-mytextbookswheniwantem-capacitance-and-dielectrics.html?page=2

Read section 26.1 mainly, beginning with “Let’s consider a capacitor formed from a pair of parallel plates as shown in Figure 26.2.”

yea that pretty much states what i was thinking. i understood how th elctrons are reppelled and attracted by the terminals which causes plates to charge i just wasnt sure if this movement in charge and change in charge on the plates affected potential difference but i see that it does. i guess when the charge moves it effect the elctric potential and pd is 0 when the elctric potntial is the same everywhere as no energy per unit charge is needed to move and essentialy it is equipotential.
Original post by charlie.R.12
yea that pretty much states what i was thinking. i understood how th elctrons are reppelled and attracted by the terminals which causes plates to charge i just wasnt sure if this movement in charge and change in charge on the plates affected potential difference but i see that it does. i guess when the charge moves it effect the elctric potential and pd is 0 when the elctric potntial is the same everywhere as no energy per unit charge is needed to move and essentialy it is equipotential.

It seems that you have understood and cleared your doubts.

A few points to note:
When the capacitor is fully charged, the potential difference across between the capacitor plates exactly matches the potential difference between the battery terminals: the battery does not really “lose” says 6 volts while the capacitor also has not really gained 6 volts. The key concept is “A potential difference or voltage is created by the separation of charges.”

“pd is 0 when the electric potential is the same everywhere as no energy per unit charge is needed to move and essentially it is equipotential.” We seldom use the term equipotential here (not that it is wrong). Instead, we would use electrostatic equilibrium. When the capacitor is fully charged, charges are no longer in motion (note that I mean net movement), the positive capacitor plate, the connecting wire, and the positive terminal of the battery form a single conductor in electrostatic equilibrium.

As you are doing A level, it would be recommended that you check your spelling and sentence structure to avoid ambiguous interpretation. This is important in A-level physics.
Original post by Eimmanuel
It seems that you have understood and cleared your doubts.

A few points to note:
When the capacitor is fully charged, the potential difference across between the capacitor plates exactly matches the potential difference between the battery terminals: the battery does not really “lose” says 6 volts while the capacitor also has not really gained 6 volts. The key concept is “A potential difference or voltage is created by the separation of charges.”

“pd is 0 when the electric potential is the same everywhere as no energy per unit charge is needed to move and essentially it is equipotential.” We seldom use the term equipotential here (not that it is wrong). Instead, we would use electrostatic equilibrium. When the capacitor is fully charged, charges are no longer in motion (note that I mean net movement), the positive capacitor plate, the connecting wire, and the positive terminal of the battery form a single conductor in electrostatic equilibrium.

As you are doing A level, it would be recommended that you check your spelling and sentence structure to avoid ambiguous interpretation. This is important in A-level physics.

cheers