# Understanding parallel plate capacitors [closed]

I need help with understanding this assignment. I originally though of using the equation of Eo*A/D = Q/ΔV but I don't know how to answer this without actual numbers to plug in besides the epsilon-not value.

The parallel plate capacitor consists of two metal plates of area A separated by a constant distance d and connected to a source of electric potential difference ΔV. One plate is connected to a high potential terminal of a voltage source (+) and the other to the low potential terminal (-). The difference in potential (the "voltage" applied to the capacitor) is the difference between the electric potentials (voltages) of the terminals. When a potential difference ΔV is applied to the plates, charges +Q and - Q build up on the + and - plates of the capacitor, respectively. Although technically always zero, the charge of the capacitor is taken to be Q.

Suppose you have a parallel plate capacitor of fixed plate area A and fixed plate spacing d. Explain what happens to the magnitude of the charge Q as the potential difference ΔV between the plates is first increased from zero to a maximum value, and then decreased from the maximum value back to zero. How does the potential change cause the charge to change.

Now suppose you have a parallel plate capacitor with a variable plate area and constant plate spacing connected to a constant source of potential difference, such as a battery. (The area in the capacitor equation is actually the area where plates overlap to face each other. Variable area can be achieved by moving the plates sideways to change the area where they overlap.) What happens to the charge Q as the plate area changes from zero to maximum value and then back to zero again by sliding one plate past the other? What happens to the charge magnitude and how does changing the area changes the charge?

Suppose you have a parallel plate capacitor of fixed plate area A connected to a source of fixed potential difference ΔV. What will happen to charge Q if the plate spacing is increased? What will happen to Q if the plate spacing is decreased? Describe what happens to Q and explain how it happens physically in response to the changing spacing.

Explain what would happen if the plates are allowed to touch each other while there are equal and opposite charges on them.

How does a dielectric filling the space between the plates cause the capacitance to increase? That is to say, what is it about the dielectric that causes the charge on the plates to increase even with no change in voltage, thus increasing the capacitance?

## closed as off-topic by CR Drost, Norbert Schuch, user36790, Kyle Kanos, ACuriousMind♦Feb 6 '16 at 14:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – CR Drost, Norbert Schuch, Community, Kyle Kanos, ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

• As per the homework policy, you will have to give us a bit more information about what you tried before we can answer. I'll say this though - you have an equation which describes capacitors, and a description of a process happening to a capacitor (area changing from 0 to max and back to zero). The equation is always true (at each area), so as the area changes, what does the equation say happens to the charge? – levitopher Feb 6 '16 at 2:40
• So, many questions in one. – Anubhav Goel Feb 6 '16 at 5:47
• Please try to ask a single, specific question in one post, not a host of more-or-less related questions. Also, don't just throw your homework assignment at us, but ask a conceptual question instead. – ACuriousMind Feb 6 '16 at 14:26

You have the right equation. Since you are not given absolute values, you should just reference your answers to the original $Q$. For example, you can say for the first part "The charge will be 0 when the voltage difference is 0. As the voltage difference increases to $\Delta V$, the charge will increase (proportionally) to $Q$". No numbers were needed...
So don't "plug in numbers" - that is using equations without understanding. This assignment is about getting a feeling for the equation. What happens to $x$ when I do $y$. Look closely at the equation, and it will give you all the answers.
Except for the last one. There, you need to adjust your equation (you have another equation in your book that contains not just $\epsilon_0$, but $\epsilon_r$...)