capacitanceelectric-circuitselectricityhomework-and-exercises
So above are two difference circuits, each with a 9V battery and a capacitor of the same capacitance. $C_1=C_2$
Why is the potential difference across both capacitors 9V? Shouldn't it be less than 9V? The potential difference across the top of the circuit and the bottom of the circuit is 9V. So why is the potential difference along a very tiny capacitor 9V too? The distance between the two capacitor plates arent the same distance as between the top of the circuit and the bottom
This is a system of two capacitors in parallel, so the total charge is given by $$ Q_\mathrm{tot} = (C_1 + C_2)V ~.$$ When a dielectric is inserted into capacitor 1, it's capacitance changes: $C_1 \to \kappa\,C_1$. However, the total charge is conserved. Thus $$ Q_\mathrm{tot} = (\kappa\,C_1 + C_2)V' ~. $$ This implies that \begin{align} (C_1 + C_2)V &= (\kappa\,C_1 + C_2)V' \\ \Rightarrow V' &= \frac{(C_1 + C_2)V}{\kappa\,C_1 + C_2} \\ \therefore V' &= \frac{Q_\mathrm{tot}}{\kappa\,C_1 + C_2} ~. \end{align} If this result is not working wherever you are submitting it, then it may be that the website is wrong, or you are misreading the question.
then why is there no potential difference between the two capacitors
It's not quite clear what you mean here but do understand that charged capacitors are electrically neutral.
When a capacitor is "charged", it is not electrically charged, it is energy charged in the same sense as when we say a battery is charged.
There is nothing mysterious about two series connected circuit elements having different voltage drops. Think of two series connected resistors with different resistor values.
I would have thought that as plate B is positively charged and plate C is negatively charged, there would be a potential difference between the two
You're forgetting something fundamental: The plates B and C along with the wire that connects them are conductors. But, for an ideal conductor, charge distributes itself so that there is no (static) potential difference across the conductor.
The voltage between the bottom plate of C1 and the top plate of C2 is zero precisely because a conductor connects the two plates.
Best Answer
For a capacitor with closer plate spacing (all else being equal), the electric field in the dielectric between the plates is stronger.
Assuming a uniform electric field, the potential difference is given by the product of the spacing and the strength of the electric field:
$$\Delta V = E\cdot d $$
So, the potential difference can be the same for both capacitors even though the plate spacing is different but the electric field between the plates will be different.
This is due to the fact that, for smaller spacing, the capacitance $C$ is larger and, thus, more charge $Q$ must be moved by the battery to charge the capacitor to 9V.
$$\Delta V = \frac{Q}{C}$$