Capacitors and Capacitance - AP Physics 2
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If the maximum amount of charge held by a capacitor at a voltage of 12V is 36C, what is the capacitance of this capacitor?
If the maximum amount of charge held by a capacitor at a voltage of 12V is 36C, what is the capacitance of this capacitor?
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In this question, we're told the maximum amount of charge that a capacitor can hold at a given voltage. We're then asked to determine the capacitance. To do this, we'll need to use the expression for capacitance.
Plug in the values given to us in the question stem:
In this question, we're told the maximum amount of charge that a capacitor can hold at a given voltage. We're then asked to determine the capacitance. To do this, we'll need to use the expression for capacitance.
Plug in the values given to us in the question stem:
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Four arrangements of capacitors are pictured. Each has an equivalent capacitance. Rank these four arrangements from highest equivalent capacitance to lowest. Assume that all capacitors are identical.
Four arrangements of capacitors are pictured. Each has an equivalent capacitance. Rank these four arrangements from highest equivalent capacitance to lowest. Assume that all capacitors are identical.
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Let's go through all of them and find the equivalent capacitance.
(A) This is just
(B) There are two capacitors in series, so this is
(C) These capacitors are in parallel, so
(D) These are a combination of series and parallel. Two are in series and they are in parallel with a third,
So, ranking them we get
Let's go through all of them and find the equivalent capacitance.
(A) This is just
(B) There are two capacitors in series, so this is
(C) These capacitors are in parallel, so
(D) These are a combination of series and parallel. Two are in series and they are in parallel with a third,
So, ranking them we get
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Consider the circuit:
What is the total equivalent capacitance?
Consider the circuit:
What is the total equivalent capacitance?
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Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:
The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:
Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:
The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:
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In this parallel-plate capacitor, the distance 
between the plates is 
, the area of each plate is 
, and the voltage across them is 
.
Calculate the capacitance.
In this parallel-plate capacitor, the distance
Calculate the capacitance.
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The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.
For a parallel-plate capacitor,
Plugging in the numbers given results in
Be careful to convert the units to meters!
The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.
For a parallel-plate capacitor,
Plugging in the numbers given results in
Be careful to convert the units to meters!
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You have 3 capacitors in series. Their capcitance's are 
, 
, and 
. What is the total capacitance of the system?
You have 3 capacitors in series. Their capcitance's are
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To find the total capacitance of capacitors in series, you use the following equation:
.
Our values for 
, 
, and 
are 4, 3, and 2. Now, we can plug in our values to find the answer.
The answer we have is the inverse of this. Therefore, the total capacitance is 
.
To find the total capacitance of capacitors in series, you use the following equation:
Our values for
The answer we have is the inverse of this. Therefore, the total capacitance is
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You have 4 capacitors, 
, 
, 
, and 
, arranged as shown in the diagram below.
Their capacitances are as follows:
What is the total capacitance of the circuit?
You have 4 capacitors,
Their capacitances are as follows:
What is the total capacitance of the circuit?
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Remember, the equations for adding capacitances are as follows:
Capacitors 
and 
are in series, 
and 
are in parallel, and 
and 
are in parallel.
Remember, the equations for adding capacitances are as follows:
Capacitors
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Three capacitors are in parallel with each other. What can be said for certain about the total capacitance?
Three capacitors are in parallel with each other. What can be said for certain about the total capacitance?
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Capacitors in parallel combine according to the following equation:
.
Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.
Capacitors in parallel combine according to the following equation:
Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.
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Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the voltage difference between the plates increase, decrease or stay the same?
Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the voltage difference between the plates increase, decrease or stay the same?
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The fact that the system is still connected to the battery indicate a constant V so regardless what happens to the capacitor, V stays fixed.
The fact that the system is still connected to the battery indicate a constant V so regardless what happens to the capacitor, V stays fixed.
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Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the charge on each plate increase, decrease or stay the same?
Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the charge on each plate increase, decrease or stay the same?
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Relevant equations:
Plug the second equation into the first:
Considering all the variables in the numerator are held fixed for this problem, we see that increasing D will decrease the charge stored on each plate.
Relevant equations:
Plug the second equation into the first:
Considering all the variables in the numerator are held fixed for this problem, we see that increasing D will decrease the charge stored on each plate.
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Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is not connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the charge on each plate increase, decrease or stay the same?
Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is not connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the charge on each plate increase, decrease or stay the same?
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The charge has no where to go. Without the battery connected, the charge has no physical avenue on or off the plates.
The charge has no where to go. Without the battery connected, the charge has no physical avenue on or off the plates.
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Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, how did the capacitance change? Also, do we need to use the information provided about the battery?
Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, how did the capacitance change? Also, do we need to use the information provided about the battery?
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As D increases, C will decrease
Does the battery matter? No. Capacitance is a "geometric" quantity. This means that it can be determined solely by physical parameters like the area and separation distance. The charge on the plates, voltage difference, electric field and any other quantity you could think of does not influence the capacitance other than A or D.
As D increases, C will decrease
Does the battery matter? No. Capacitance is a "geometric" quantity. This means that it can be determined solely by physical parameters like the area and separation distance. The charge on the plates, voltage difference, electric field and any other quantity you could think of does not influence the capacitance other than A or D.
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Consider the given diagram. If 
, each plate of the capacitor has surface area 
, and the plates at 
apart, determine the excess charge on the positive plate.
Consider the given diagram. If
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The voltage rise through the source must be the same as the drop through the capacitor.
The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.
Combine equations and solve for the electric field:
Convert mm to m and plug in values:
Using the electric field in a capacitor equation:
Rearrange to solve for the charge:
Convert 
to 
and plug in values:
The voltage rise through the source must be the same as the drop through the capacitor.
The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.
Combine equations and solve for the electric field:
Convert mm to m and plug in values:
Using the electric field in a capacitor equation:
Rearrange to solve for the charge:
Convert
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Consider the given diagram. If 
, each plate of the capacitor has surface area 
, and the plates are 0.1mm apart, determine the number of excess electrons on the negative plate.
Consider the given diagram. If
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The voltage rise through the source must be the same as the drop through the capacitor.
The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.
Combine equations and find the electric field:
Convert mm to m and plugg in values:
Use the electric field in a capacitor equation:
Convert 
to 
and plug in values:
The magnitude of total charge on the positive plate is equal to the total charge on the negative plate, so to find the number of excess elections:
The voltage rise through the source must be the same as the drop through the capacitor.
The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.
Combine equations and find the electric field:
Convert mm to m and plugg in values:
Use the electric field in a capacitor equation:
Convert
The magnitude of total charge on the positive plate is equal to the total charge on the negative plate, so to find the number of excess elections:
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Suppose I have a uniform electric field within a parallel plate capacitor with field strength of 
.
Suppose the capacitor's plates are 
in length and 
in width, and the space between the plates is 
.
Determine the capacitance of this system given that the space in between is a vacuum, and that the permittivity of empty space 
is 
.
Suppose I have a uniform electric field within a parallel plate capacitor with field strength of
Suppose the capacitor's plates are
Determine the capacitance of this system given that the space in between is a vacuum, and that the permittivity of empty space
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The formula for capacitance 
is given by:
,
Where 
is dielectric strength, 
is distance between plates, 
is permittivity of empty space, and 
is cross sectional area.
To determine 
, we do
because between the plates is a vacuum.
Putting it all together,
The formula for capacitance
Where
To determine
Putting it all together,
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Suppose I have a uniform electric field within a parallel plate capacitor with field strength of 
.
Suppose the capacitor's plates are 
in length and 
in width, and the space between the plates is 
.
What is the voltage difference across the capacitor?
Suppose I have a uniform electric field within a parallel plate capacitor with field strength of
Suppose the capacitor's plates are
What is the voltage difference across the capacitor?
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Voltage difference 
is given by
, where 
is the electric field strength and 
is the distance between the two plates.
For this problem,
Voltage difference
For this problem,
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Consider the circuit:
What is the total equivalent capacitance?
Consider the circuit:
What is the total equivalent capacitance?
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Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:
The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:
Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:
The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:
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In this parallel-plate capacitor, the distance between the plates is , the area of each plate is , and the voltage across them is .
Calculate the capacitance.
In this parallel-plate capacitor, the distance
Calculate the capacitance.
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The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.
For a parallel-plate capacitor,
Plugging in the numbers given results in
Be careful to convert the units to meters!
The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.
For a parallel-plate capacitor,
Plugging in the numbers given results in
Be careful to convert the units to meters!
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You have 3 capacitors in series. Their capcitance's are , , and . What is the total capacitance of the system?
You have 3 capacitors in series. Their capcitance's are
Tap to reveal answer
To find the total capacitance of capacitors in series, you use the following equation:
.
Our values for , , and are 4, 3, and 2. Now, we can plug in our values to find the answer.
The answer we have is the inverse of this. Therefore, the total capacitance is .
To find the total capacitance of capacitors in series, you use the following equation:
Our values for
The answer we have is the inverse of this. Therefore, the total capacitance is
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You have 4 capacitors, , , , and , arranged as shown in the diagram below.
Their capacitances are as follows:
What is the total capacitance of the circuit?
You have 4 capacitors,
Their capacitances are as follows:
What is the total capacitance of the circuit?
Tap to reveal answer
Remember, the equations for adding capacitances are as follows:
Capacitors and are in series, and are in parallel, and and are in parallel.
Remember, the equations for adding capacitances are as follows:
Capacitors
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Three capacitors are in parallel with each other. What can be said for certain about the total capacitance?
Three capacitors are in parallel with each other. What can be said for certain about the total capacitance?
Tap to reveal answer
Capacitors in parallel combine according to the following equation:
.
Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.
Capacitors in parallel combine according to the following equation:
Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.
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