All AP Physics C: Mechanics Resources
Example Questions
Example Question #1 : Electric Circuits
A parallel plate capacitor has a capacitance of . If the plates are
apart, what is the area of the plates?
The relationship between capacitance, distance, and area is . We can rearrange this equation to solve for area.
Now, we can use the values given in teh question to solve.
Example Question #1 : Capacitors
Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area
and are separated by distance
. A material of dielectric constant
has been placed between the two plates.
Which of the following would not result in an increase in the measure of electric potential difference between the two plates?
Increase the area, , of the plates
Replace the material between the two plates with one of a lower dielectric constant,
Increase the value of the surface charge density, , on each plate
Increase the distance, , between the plates
Increase the area, , of the plates
The electric potential difference created between the plates of a parallel plate capactor is given by the equation:
The charge can be calculated by using the equation:
The value of the capacitance is related to the dimensions of the capacitor with the equation:
Combining these equations yields:
The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.
Example Question #1 : Using Capacitor Equations
You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of out of those metal sheets, and the sheets needed to be
apart, what area do the two metal plates need to be?
For parallel plate capacitors, the equation is .
Solve for .
Now we can plug in our given values.
Example Question #1 : Electric Circuits
The plates of a parallel plate capacitor are apart and
in area. A potential difference of
is applied across the capacitor. Find the capacitance.
Capacitance is related to plate area and distance by the equation .
Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.
Example Question #1 : Electric Circuits
The plates of a parallel plate capacitor are apart and
in area. A potential difference of
is applied across the capacitor. Compute the charge on each plate.
Charge on a capacitor is given by the equation . We know that the voltage is
, but we need to determine the capacitance based off of the area and distance between the plates.
We can plug this value into the equation for charge.
Example Question #1 : Capacitors
The plates of a parallel plate capacitor are apart. A potential difference of
is applied across the capacitor. Compute the magnitude of the generated electric field.
The electric field given by a capacitor is given by the formula . We do not have these variables, so we will have to adjust the equation.
The capacitance can be determined by the area of the plates and the distance between them.
This equation simplifies to , allowing us to solve using the values given in the question.
Example Question #1 : Electric Circuits
If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance,
?
Relevant equations:
Use the series equation, replacing C1 and C2 with the given constant C:
This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.
Example Question #1 : Using Capacitor Equations
Two parallel conducting plates each have an area of and are separated by a distance,
. One plate has a charge
evenly distributed across it, and the other has a charge of
. A proton (charge
) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?
Relevant equations:
According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.
1. Find an expression for potential difference, , in terms of
and
, by setting the two capacitance equations equal to each other:
2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):
Example Question #211 : Ap Physics C
A and
capacitor are connected in series with a
battery. What is the magnitude of the charge on one plate of the
capacitor?
Relevant equations:
First, find the equivalent capacitance of the whole circuit:
Use this equivalent capacitance to find the total charge:
Capacitors in series always have the same charge on each unit, so the charge on the capacitor is also
.
All AP Physics C: Mechanics Resources
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