Capacitors and Potential Difference

In summary: By using the given values for q and v, we can find the potential at a point midway between the plates. For the second point, we can use the formula PE = 1/2CV^2, where C is the capacitance between the point and the plate, and V is the potential difference. By substituting the given values for C and V, we can find the potential at the second point. Subtracting these two values will give us the potential difference between the two points, which is the desired answer. In summary, we can use the equations for electric potential and electric potential energy to find the potential difference between two points in a parallel plate capacitor.
  • #1
Kaoi
21
0

Homework Statement


(Note: This is the fourth part of a single question, but it's the only part I'm having trouble with. Don't worry, it's not as simple as Q/V. :rofl:)

"What is the potential difference between a point midway between the [circular, parallel] plates [which are 1.43 x 10-4 m apart] and a point that is 1.23 x 10-4 m from one of the plates? Answer in units of V."

Givens: (Some of these I have gotten from the (proven correct) solutions from other parts of the question.)

[tex]\Delta V_{0} = 0.148 V[/tex]
[tex]\Delta d_{1} = 7.15 \times 10^{-5} m[/tex]
[tex]\Delta d_{2} = 1.23 \times 10^{-4} m[/tex]
[tex]C_{0} = 5.264 \times 10^{-15} F[/tex]
[tex]Q_{0} = 7.79 \times 10^{-14} C[/tex]

Needed:
[tex]\Delta V_{N} = ?[/tex]

Homework Equations


[tex]C = \frac{Q}{\Delta V}[/tex]
[tex]\Delta V = E\Delta d = \frac{\Delta PE_{e}}{q}[/tex]
[tex]PE_{electric} = \frac{1}{2}C\Delta V^{2} = \frac{Q^{2}}{2C}[/tex]

The Attempt at a Solution



Alright. My problem with this question isn't so much mathematical as conceptual. If I could figure out some things, I could definitely apply some equations to solve it.

-Should I use point-charge equations for this?
-Since we're talking about voltage between points and not charges, can I use the charge capacity of the capacitor in my equations?
-Can I add up the distances between the points, or do I need to work them separately because they are on different sides of the source of the field?
-If I change the voltage, wouldn't I be unable to use my values for [tex]C[/tex] and [tex]Q[/tex], since they rely on a certain voltage?
 
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  • #2
Since in a parallel plate capacitor, the electric field is constant at all points, the potential gradient must also be a constant. Therefore, the potential at a point midway will be q/2v.
 
  • #3


As a scientist, it is important to first understand the concepts involved before attempting to solve a problem. Let's break down the given information and equations to understand the problem better.

1. Capacitors: A capacitor is a device that stores electrical energy in the form of an electric field. It consists of two parallel plates separated by a distance, with a dielectric material in between. The capacitance of a capacitor is the measure of its ability to store charge.

2. Potential Difference: Potential difference, also known as voltage, is the difference in electric potential between two points in an electric field. It is measured in volts (V).

3. Distance between plates: In this problem, we are given the distance between the parallel plates, which is 1.43 x 10^-4 m. This distance is important because it affects the capacitance of the capacitor.

4. Charge and voltage: We are also given the charge (Q) and voltage (ΔV) values, which are important in calculating the potential difference between the two points.

Now let's look at the given equations:

1. C = Q/ΔV: This equation relates the capacitance of a capacitor to the charge and voltage. It tells us that the capacitance is directly proportional to the charge and inversely proportional to the voltage.

2. ΔV = EΔd = ΔPEe/q: This equation relates the potential difference between two points to the electric field and the distance between the points. It tells us that the potential difference is directly proportional to the electric field and the distance between the points.

3. PEelectric = 1/2CΔV^2 = Q^2/2C: These equations relate the potential energy of a capacitor to its capacitance and voltage. They tell us that the potential energy is directly proportional to the capacitance and the square of the voltage.

Now, let's apply these concepts and equations to solve the problem.

1. Since the problem asks for the potential difference between a point midway between the plates and a point 1.23 x 10^-4 m from one of the plates, we need to consider the electric field created by the capacitor. This means we can use the equation ΔV = EΔd to calculate the potential difference.

2. We are given the distance between the plates (1.43 x 10^-4 m) and the distance from one of the plates (1.23 x
 

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy by accumulating an electric charge on its plates, separated by an insulating material called a dielectric. It is commonly used in circuitry to store energy and regulate the flow of electricity.

2. How do capacitors work?

Capacitors work by accumulating an electric charge on its plates, which creates an electric field between them. The charge is stored in the form of an electric potential difference, or voltage, between the plates. When connected to a circuit, the capacitor allows the flow of current until it reaches its maximum capacity, at which point it blocks any further current.

3. What is the role of capacitors in circuits?

Capacitors have various roles in circuits, such as smoothing out fluctuations in voltage, blocking direct current while allowing alternating current to pass, and storing electrical energy. They can also be used to filter out unwanted frequencies, act as timing devices, and even boost signals.

4. How is potential difference related to capacitors?

Potential difference, also known as voltage, is directly related to capacitors as it is the measure of the electric potential energy stored in the capacitor. The larger the potential difference, the more energy the capacitor can store. This relationship is described by the equation Q=CV, where Q is the charge, C is the capacitance, and V is the potential difference.

5. What factors affect the capacitance and potential difference of a capacitor?

The capacitance of a capacitor is affected by the distance between its plates, the surface area of the plates, and the type of dielectric used. The potential difference, or voltage, is determined by the amount of charge stored and the capacitance. It can be increased by increasing the amount of charge or by decreasing the capacitance.

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