$V_0−\frac QC=IR$? Is this what is happening in this equation, or is it at least close?

Yes, naturally, this is just a rearranging of the equation you have set up. Nothing is wrong with this.

But you have to distinguish current from voltage.

• There is a voltage (a potential difference) from the positive to the negative terminal of the battery. This voltage is the "push" that tries to push charges around the circuit. It "pushes" constantly - but that doesn't mean that any charges necessarily are moving...

• The current in this circuit changes until the capacitor is fully charge - after which it is 0! No current can flow if there is a hole in the circuit - a capacitor is a hole in a circuit. But this is only the case when the capacitor is fully charged, because...

• Initially, the current "doesn't know" that there is a hole. Electrons flow from one capacitor plate towards the battery's positive terminal, and electrons flow from the negative terminal towards the other capacitor plate. They move (current flows) as if the circuit is closed.

• Pretty soon the reach the end and can't move further. Electrons gather up at the lower capacitor plate and acculumate here. The charge $-Q$ built up on this lower plate, induced the exact same charge $+Q$ of opposite sign on the other plate, because they are so close. They together set up a stronger and stronger counter-working electric field.

• At some point this counter-working electric field repels incoming electrons just as much as the negative terminal repels them. There is no net force on charges anymore and all charge-flow stops. This situation now looks like an open circuit.

Conclusion is that after some time (usually a very short time, but that depends on the capacitor) no more current is flowing. The equation you have shown is correct at only moment in time; just keep in mind that the values of $I$ and $Q$ change constantly (one increases and one decreases) during the charging of the capacitor. And when fully charged, $I=0$ and $Q$ is at it's maximum.

$V_0−\frac QC=IR$? Is this what is happening in this equation, or is it at least close?

Yes, naturally, this is just a rearranging of the equation you have set up. Nothing is wrong with this.

But you have to distinguish current from voltage.

• There is a voltage (a potential difference) from the positive to the negative terminal of the battery. This voltage is the "push" that tries to push charges around the circuit. It "pushes" constantly - but that doesn't mean that any charges necessarily are moving...

• The current in this circuit changes until the capacitor is fully charge - after which it is 0! No current can flow if there is a hole in the circuit - a capacitor is a hole in a circuit. But this is only the case when the capacitor is fully charged, because...

• Initially, the current "doesn't know" that there is a hole. Electrons flow from one capacitor plate towards the battery's positive terminal, and electrons flow from the negative terminal towards the other capacitor plate. They move (current flows) as if the circuit is closed.

• Pretty soon they reach the end and can't move further. Electrons gather up at the lower capacitor plate and acculumate here. The charge $-Q$ built up on this lower plate, induced the exact same charge $+Q$ of opposite sign on the other plate, because they are so close. The accumulating charges set up a counter-working electric field, growing stronger and stronger as more charges arrive.

• At some point this counter-working electric field repels incoming electrons just as much as the negative terminal repels them. (Vice versa for the built up positive charge on the other plate attracting electrons just as much as the positive terminal does.) There is no net force on charges anymore and all charge-flow stops. This situation now looks like an open circuit.

Conclusion is that after some time (usually a very short time, but that depends on the capacitor) no more current is flowing. The equation you have shown is correct at only moment in time; just keep in mind that the values of $I$ and $Q$ change constantly (one increases and one decreases) during the charging of the capacitor. And when fully charged, $I=0$ and $Q$ is at it's maximum.

$V_0−\frac QC=IR$? Is this what is happening in this equation, or is it at least close?

Yes, naturally, this is just a rearranging of the equation you have set up. Nothing is wrong with this.

But you have to distinguish current from voltage.

• There is a voltage from the positive to the negative terminal of the battery. In other words, a voltage is the "push" that tries to push current around the circuit. But that doesn't mean that current is moving...

• Current in this circuit changing until the capacitor is fully charge - after which is is 0! No current can flow if there is a hole in the circuit - a capacitor is a hole in a circuit. But this is only the case when the capacitor is fully charged, because...

• Initially, the current "doesn't know" that there is a hole. Electrons flow from one capacitor plate towards the battery's positive terminal, and electrons flow from the negative terminal towards the other capacitor plate. They move (current flows) as if the circuit is closed.

• Pretty soon the reach the end and can't move further. Electrons gather up at the lower capacitor plate and acculumate here. The charge $-Q$ built up on this lower plate, induced the exact same charge $+Q$ of opposite sign on the other plate, because they are so close. They together set up a stronger and stronger counter-working electric field.

• At some point this counter-working electric field repels incoming electrons just as much as the negative terminal repels them. There is no net force on charges anymore and all charge-flow stops. This situation now looks like an open circuit.

Conclusion is that after some time (usually a very short time, but that depends on the capacitor) no more current is flowing. The equation you have shown is correct at only moment in time; just keep in mind that the values of $I$ and $Q$ change (one increases and one decreases) constantly during the charging of the capacitor. And when fully charged, $I=0$ and $Q$ is at it's maximum.

$V_0−\frac QC=IR$? Is this what is happening in this equation, or is it at least close?

Yes, naturally, this is just a rearranging of the equation you have set up. Nothing is wrong with this.

But you have to distinguish current from voltage.

• There is a voltage (a potential difference) from the positive to the negative terminal of the battery. This voltage is the "push" that tries to push charges around the circuit. It "pushes" constantly - but that doesn't mean that any charges necessarily are moving...

• The current in this circuit changes until the capacitor is fully charge - after which it is 0! No current can flow if there is a hole in the circuit - a capacitor is a hole in a circuit. But this is only the case when the capacitor is fully charged, because...

• Initially, the current "doesn't know" that there is a hole. Electrons flow from one capacitor plate towards the battery's positive terminal, and electrons flow from the negative terminal towards the other capacitor plate. They move (current flows) as if the circuit is closed.

• Pretty soon the reach the end and can't move further. Electrons gather up at the lower capacitor plate and acculumate here. The charge $-Q$ built up on this lower plate, induced the exact same charge $+Q$ of opposite sign on the other plate, because they are so close. They together set up a stronger and stronger counter-working electric field.

• At some point this counter-working electric field repels incoming electrons just as much as the negative terminal repels them. There is no net force on charges anymore and all charge-flow stops. This situation now looks like an open circuit.

Conclusion is that after some time (usually a very short time, but that depends on the capacitor) no more current is flowing. The equation you have shown is correct at only moment in time; just keep in mind that the values of $I$ and $Q$ change constantly (one increases and one decreases) during the charging of the capacitor. And when fully charged, $I=0$ and $Q$ is at it's maximum.