This activity uses a plastic surface, like the one in the figure, to represent the electric potential between two charged plates, one positive and one negative. A \(1 \mathrm{~cm}\) height difference corresponds to an electric potential difference of \(1 \mathrm{~V}\text{.}\)
Align your surface with a contour map. How are you making this alignment? Where is the surface a good approximation of the potential? Where is the approximation less good?
Propose an equation to describe the electric potential as a function of the distance, \(x\text{,}\) from the negative plate. Where did choose for the location of \(V = 0\text{?}\)
The electric field inside a parallel plate capacitor is uniform with magnitude \(|E| = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}\text{,}\) where \(Q\) is the total charge on the positive plate, \(A\) is the area of each plate, and \(d\) is the distance between the plates.
How is \(\Delta V\) related to \(Q\text{?}\) Sketch a graph with \(Q\) on the vertical axis and \(\Delta V\) on the horizontal axis showing the relationship between them.
This slope is known as the capacitance. Capacitance can be calculated as \(C = Q/\Delta V\) and represents the amount of charge needed on each plate per volt of potential difference between the plates. Capacitance is a constant that is a characteristic of a given capacitor and is measured in farads (F).
Consider a change to the setup of the capacitor: the area of the capacitor plates increases, while the total charge on each plate and the distance between the plates remain the same.
How does \(\Delta V\) between the two plates compare to the original situation? Give an answer based on an equation and give an explanation based on the physical situation.
Consider a different change to the setup of the original capacitor: the distance between the capacitor plates increases, while the total charge on each plate and the area of the plates remain the same.
How does \(\Delta V\) between the two plates compare to the original situation? Give an answer based on an equation and give an explanation based on the physical situation.
Use the battery to give the capacitor some initial charge, then disconnect the battery from the capacitor. Why do you think disconnecting the battery from the capacitor will hold the charge constant?
Connect the capacitor to the battery again. Adjust \(A\) and \(d\) for the capacitor and observe the changes. What is held constant when you adjust \(A\) and \(d\text{?}\) How are the changes similar to what happened when the capacitor was disconnected from the battery? How are the changes different?
A uniformly charged ring with total charge \(-q_o\) and radius \(R\) is located in the \(xy\)-plane. You previously found the electric potential a distance \(z\) above the center of the ring: \(V(z)=-\frac{kq_o}{\sqrt{R^2+z^2}}\text{.}\)
Sketch and label a graph of the given electric potential vs. position \(z\text{.}\) You may use a graphing calculator to help construct the shape of your graph.
Sketch and label a graph of the \(z\)-component of the electric field vs. position. Use the graph to add electric field vectors to your diagram from part A.
What direction do you expect the electric field to point when \(z \gt 0\text{?}\) When \(z \lt 0\text{?}\) Check that your equations match this expectation.
What do you expect for the value of the potential at \(z = 0\text{?}\) What about the electric field? Check that your equations match this expectation.
What do you expect for the functional form of the potential as \(z\) becomes much larger than \(R\text{?}\) What about the electric field? Check that your equations match this expectation.
