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?