20.1 Electric Potential and Potential Difference Raymond A. Serway John W. Jewett
•
The potential energy per unit charge, U/qo, is the electric potential • •
Chapter 20
•
The potential is independent of the value of qo The potential has a value at every point in an electric field
The electric potential is
Electric Potential and Capacitance •
20.1 Electric Potential and Potential Difference •
When a test charge, qo, is placed in an electric field, it experiences a force •
•
The force is conservative
20.1 Electric Potential and Potential Difference •
•
As a charged particle moves in an electric field, it will experience a change in potential The potential difference is
To find the work associated with the force, look at an infinitesimal displacement vector that is oriented tangent to a path through space This path may be straight or curved, and an integral performed along this path is called either a path integral or a line integral
•
•
•
20.1 Electric Potential and Potential Difference •
The work done by the electric field is •
As this work is done by the field, the potential energy of the charge-field system is changed by U
For a finite displacement of the charge from A to B, •
Just as with potential energy, only differences in electric potential are meaningful We often take the value of the potential to be zero at some convenient point in the field
20.1 Electric Potential and Potential Difference •
The potential is characteristic of the field only •
•
Electric potential is a scalar
•
•
It is independent of the charge placed in the field The difference in potential is proportional to the difference in potential energy
Potential energy is characteristic of the charge-field system •
Due to an interaction between the field and a charged particle placed in the field
Because the force is conservative, the line integral does not depend on the path taken by the charge
1
20.1 Electric Potential and Potential Difference •
The electric potential at an arbitrary point due to source charges equals the work required by an external agent to bring a test charge from infinity to that point divided by the charge on the test particle •
•
SI units of both electric potential and potential difference is the volt: 1 V = 1 J/C It takes one Joule of work to move a 1 Coulomb charge through a potential difference of 1 Volt
Because E is constant:
20.2 Potential Difference in a Uniform Electric Field •
•
The negative sign indicates that the electric potential at B is lower than at A VB < VA Electric field lines always point in the direction of decreasing electric potential
We can interpret the electric field as a measure of the rate of change with position of the electric potential
20.1 Electric Potential and Potential Difference
•
Consider a uniform electric field directed along the negative y axis The potential difference between the two points is:
•
The SI units of the electric field are: 1 N/C = 1 V/m •
•
•
The work performed on the charge is
•
•
•
Assumes a charge moves in an electric field without any change in its kinetic energy
20.1 Electric Potential and Potential Difference •
20.2 Potential Difference in a Uniform Electric Field
Another unit of energy that is commonly used in atomic and nuclear physics is the electronvolt One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e is moved through a potential difference of 1 volt 1 eV = 1.60 x 1019 C·V = 1.60 x 1019 J
20.2 Potential Difference in a Uniform Electric Field •
•
Suppose a test charge moves from A to B The change in the potential energy of the charge-field system is:
2
20.2 Potential Difference in a Uniform Electric Field •
•
In a system consisting of a positive charge and an electric field, the electric potential energy of the system decreases when the charge moves in the direction of the field Equivalently, an electric field does work on a positive charge when the charge moves in the direction of the electric field
20.2 Potential Difference in a Uniform Electric Field •
Consider a more general case: •
•
•
20.2 Potential Difference in a Uniform Electric Field •
A system consisting of a positive charge and an electric field loses electric potential energy when the charge moves in the direction of the field •
•
An electric field does work on a positive charge when the charge moves in the direction of the electric field
The charged particle gains kinetic energy equal to the potential energy lost by the charge-field system •
Another example of conservation of energy
20.2 Potential Difference in a Uniform Electric Field • •
If qo is negative, then U is positive A system consisting of a negative charge and an electric field gains potential energy when the charge moves in the direction of the field •
In order for a negative charge to move in the direction of the field, an external agent must do positive work on the charge
A charged particle moves between A and B The displacement vector is not parallel to the field
Change in potential energy is:
20.2 Potential Difference in a Uniform Electric Field •
•
All points in a plane perpendicular to a uniform electric field are at the same electric potential The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential •
The equipotential surfaces associated with a uniform electric field consist of a family of parallel planes that are all perpendicular to the field
Example 20.1 The Electric Field Between Two Parallel Plates of Opposite Charge A battery has a specified potential difference V between its terminals and establishes that potential difference between conductors attached to the terminals. A 12-V battery is connected between two parallel plates. The separation between the plates is d = 0.30 cm, and we assume the electric field between the plates to be uniform.
3
Example 20.1 The Electric Field Between Two Parallel Plates of Opposite Charge
Example 20.2 Motion of a Proton in a Uniform Electric Field
This assumption is reasonable if the plate separation is small relative to the plate dimensions and we do not consider locations near the plate edges. Find the magnitude of the electric field between the plates. •
•
Solve for the final speed of the proton:
•
Substitute numerical values:
Evaluate the magnitude of the electric field between the plates:
Example 20.2 Motion of a Proton in a Uniform Electric Field A proton is released from rest at point in a uniform electric field that has a magnitude of 8.0×104 V/m. The proton undergoes a displacement of magnitude d = 0.50 m to point in the direction of the electric field. Find the speed of the proton after completing the displacement.
Example 20.2 Motion of a Proton in a Uniform Electric Field •
Find the potential difference:
•
Use conservation of energy:
20.3 Electric Potential and Potential Energy Due to Point Charges •
An isolated positive point charge q produces an electric field directed radially outward from the charge •
20.3 Electric Potential and Potential Energy Due to Point Charges •
•
•
To find the electric potential at a point located a distance r from the charge, begin with the general expression:
At any point in space, the field due to the point charge is:
Then we have:
Substitute: •
The dot product is: ds cos
4
20.3 Electric Potential and Potential Energy Due to Point Charges •
The expression for potential difference is:
20.3 Electric Potential and Potential Energy Due to Point Charges •
The energy appears in the system as potential energy U when the particles are separated by a distance r12 •
•
•
The integral is independent of the path
The electric force is conservative, and the field is a conservative field
20.3 Electric Potential and Potential Energy Due to Point Charges •
•
•
It is customary to choose a reference potential of V = 0 at rA = Then the potential at some point r is
•
20.3 Electric Potential and Potential Energy Due to Point Charges •
If the charges are of the same sign, then U is positive •
Positive work must be done by an external agent on the system to bring the two charges near each other (because charges of the same sign repel)
For a group of point charges, the total electric potential at P is:
20.3 Electric Potential and Potential Energy Due to Point Charges •
The potential energy of the system can be expressed as
Consider the potential energy of a system of two charged particles If V2 is the electric potential at a point P due to charge q2, the work an external agent must do to bring a second charge q1 from infinity to P without acceleration is q1V2 •
This work represents a transfer of energy into the system
20.3 Electric Potential and Potential Energy Due to Point Charges •
If the charges are of opposite sign, then U is negative •
•
Negative work is done by an external agent against the attractive force between the charges of opposite sign as they are brought near each other A force must be applied opposite the displacement to prevent q1 from accelerating toward q2
5
20.3 Electric Potential and Potential Energy Due to Point Charges •
Let’s remove charge q1 •
A potential due to charge q2 can be defined:
20.3 Electric Potential and Potential Energy Due to Point Charges •
If there are more than two charges, then find U for each pair of charges and add them
•
The total electric potential energy of a system of point charges is equal to the work required to bring the charges, one at a time, from an infinite separate to their final positions
•
Add them algebraically
Example 20.3 The Electric Potential Due to Two Point Charges A charge q1 = 2.00 C is located at the origin and a charge q2 = -6.00 C is located at (0, 3.00) m.
Example 20.3 The Electric Potential Due to Two Point Charges (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. •
Find the potential:
•
Substitute numerical values:
Example 20.3 The Electric Potential Due to Two Point Charges (B) Find the change in potential energy of the system of two charges plus a third charge q3 = 3.00 C as the latter charge moves from infinity to point P.
Example 20.3 The Electric Potential Due to Two Point Charges •
•
Evaluate the potential energy for the configuration in which the charge is at P:
Substitute numerical values to evaluate U:
6
20.4 Obtaining the Value of the Electric Field from the Electric Potential • We can calculate the value of the electric field if the electric potential is known in a region • Express the potential difference dV between two points a distance ds apart
•
The equipotential surfaces are always perpendicular to the electric field lines passing through them
20.4 Obtaining the Value of the Electric Field from the Electric Potential •
•
An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a. The dipole is along the x axis and is centered at the origin.
Assume the field has only one component Ex:
20.4 Obtaining the Value of the Electric Field from the Electric Potential •
Example 20.4 The Electric Potential Due to a Dipole
If the charge distribution creating the field has spherical symmetry, the electric field is radial, and
Example 20.4 The Electric Potential Due to a Dipole (A) Calculate the electric potential at point P on the y axis •
Find the electric potential at P due to the two charges:
Example 20.4 The Electric Potential Due to a Dipole (B) Calculate the electric potential at point R on the positive x axis •
Find the electric potential at R due to the two charges:
In general, the electric potential is a function of all three dimensions •
Given V(x, y, z) you can find Ex, Ey, and Ez as partial derivatives
7
Example 20.4 The Electric Potential Due to a Dipole
20.5 Electric Potential Due to Continuous Charge Distributions
(C) Calculate V and Ex at a point on the x axis far from the dipole
•
•
For point R far from the dipole such that x >> a, neglect a2 in the denominator of the answer to part (B) and write V in this limit:
To find the total potential, you need to integrate to include the contributions from all the elements:
•
Example 20.4 The Electric Potential Due to a Dipole •
Calculate the x component of the electric field at a point on the x axis far from the dipole:
20.5 Electric Potential Due to Continuous Charge Distributions •
Consider a small charge element dq •
•
Treat it as a point charge
The potential at some point due to this charge element is
This value for V uses the reference of V = 0 when P is infinitely far away from the charge distributions
20.5 Electric Potential Due to Continuous Charge Distributions •
If the charge distribution has sufficient symmetry, we first evaluate the field using Gauss’s law and then substitute the value obtained into
•
We then choose the electric potential V to be zero at some convenient point
Problem Solving Strategy: Calculating Electric Potential 1. Conceptualize -Think carefully about the individual charges or the charge distribution you have in the problem and imagine what type of potential would be created •
Appeal to any symmetry in the arrangement of charges to help you visualize the potential
2. Categorize - Are you analyzing a group of individual charges or a continuous charge distribution? •
The answer to this question will tell you how to proceed in the Analzye step
8
Problem Solving Strategy: Calculating Electric Potential
Problem Solving Strategy: Calculating Electric Potential
3. Analyze - When working problems involving electric potential, remember that it is a scalar quantity, so there are no components to consider
3. Analyze, cont. (a) If you are analyzing a group of individual charges:
•
•
Therefore, when using the superposition principle to evaluate the electric potential at a point, simply take the algebraic sum of the potentials due to each charge
Use the superposition principle, which states that when several point charges are present, the resultant potential at a point P in space is the algebraic sum of the individual potentials at P due to the individual charges:
Problem Solving Strategy: Calculating Electric Potential
Problem Solving Strategy: Calculating Electric Potential
3. Analyze, cont. As with potential energy in mechanics, only changes in electric potential are significant;
3. Analyze, cont. (b) If you are analyzing a continuous charge distribution:
•
hence, the point where the potential is set at zero is arbitrary
•
•
Replace the sums for evaluating the total potential at some point P from individual charges by integrals:
The charge distribution is divided into infinitesimal elements of charge dq located at a distance r from the point P
Problem Solving Strategy: Calculating Electric Potential
Problem Solving Strategy: Calculating Electric Potential
3. Analyze, cont. When dealing with point charges or a finitesized charge distribution, we usually define V = 0 to be at a point infinitely far from the charges
3. Analyze, cont. An element is then treated as a point charge, so the potential at P due to the element is dV = ke dq/r
•
If the charge distribution itself extends to infinity, however, some other nearby point must be selected as the reference point
•
•
•
The total potential at P is obtained by integrating over the entire charge distribution For many problems, it is possible in performing the integration to express dq and r in terms of a single variable To simplify the integration, consider the geometry involved in the problem
9
Problem Solving Strategy: Calculating Electric Potential
Example 20.5 Electric Potential Due to a Uniformly Charged Ring
3. Analyze, cont. To obtain the potential from the electric field:
(B) Find an expression for the magnitude of the electric field at point P
•
•
Another method used to obtain the potential is to start with the definition of the potential difference given by:
•
The electric field along can only have an x component:
If the electric field is known or can be obtained easily (such as from Gauss’s law), the line integral can be evaluated
Problem Solving Strategy: Calculating Electric Potential
Example 20.6 Electric Potential Due to a Uniformly Charged Disk
4. Finalize - Check to see if your expression for the potential is consistent with the mental representation and reflects any symmetry you noted previously
A uniformly charged disk has radius R and surface charge density .
•
Imagine varying parameters such as the distance of the observation point from the charges or the radius of any circular objects to see if the mathematical result changes in a reasonable way
Example 20.5 Electric Potential Due to a Uniformly Charged Ring
Example 20.6 Electric Potential Due to a Uniformly Charged Disk
(A) Find an expression for the electric potential at a point P located on the perpendicular central axis of a uniformly charged ring of radius a and total charge Q
(A) Find the electric potential at a point P along the perpendicular central axis of the disk
•
•
Express V in terms of the geometry: •
•
Find the amount of charge dq on a ring of radius r and width dr:
Find the potential due to the ring:
Integrate over the ring:
10
Example 20.6 Electric Potential Due to a Uniformly Charged Disk •
Integrate over the limits r = 0 to r = R:
20.6 Electric Potential Due to a Charged Conductor •
V is constant everywhere on the surface of a charged conductor in equilibrium
•
The surface of any charged conductor in electrostatic equilibrium is an equipotential surface
•
•
Evaluate the integral:
•
V = 0 between any two points on the surface
Because the electric field is zero inside the conductor, the electric potential is constant everywhere inside the conductor and equal to the value at the surface
Example 20.6 Electric Potential Due to a Uniformly Charged Disk
20.6 Electric Potential Due to a Charged Conductor
(B) Find the x component of the electric field at a point P along the perpendicular central axis of the disk
•
•
•
Find the electric field at any axial point: •
•
20.6 Electric Potential Due to a Charged Conductor •
Consider two points on the surface of the charged conductor as shown •
•
Radius is R, total charge is Q
The electric potential is a function of r The electric field is a function of r2
20.6 Electric Potential Due to a Charged Conductor •
Consider a nonspherical conductor •
The electric field is always perpendicular to the displacement: The potential difference between A and B is also zero:
Consider a solid metal conducting sphere
•
•
The surface charge density is high when the radius of curvature is small It is low when the radius of curvature is large
The electric field is large near convex points having small radii of curvature
11
Example 20.7 Two Connected Charged Spheres
20.6 Electric Potential Due to a Charged Conductor
Two spherical conductors of radii r1 and r2 are separated by a distance much greater than the radius of either sphere. The spheres are connected by a conducting wire. The charges on the spheres in equilibrium are q1 and q2, respectively, and they are uniformly charged.
•
Suppose a conductor of arbitrary shape contains a cavity
•
In this case, the electric field inside the cavity must be zero regardless of the charge distribution on the outside surface of the conductor
Example 20.7 Two Connected Charged Spheres
20.6 Electric Potential Due to a Charged Conductor
Find the ratio of the magnitudes of the electric fields at the surfaces of the spheres.
•
•
•
•
•
•
Example 20.7 Two Connected Charged Spheres Write expressions for the magnitudes of the electric fields at the surfaces of the spheres:
•
Evaluate the ratio of these two fields:
The integral must be zero since V = 0. Only true if the field is zero everywhere in the cavity
20.7 Capacitance •
•
•
Evaluate the potential difference:
Solve for the ratio of charges on the spheres: •
•
Any two points on the cavity’s surface must be at the same potential
Imagine an electric field exists in the cavity •
•
The field in the cavity is zero even if an electric field exists outside the conductor
Every point on the conductor is at the same electric potential •
Set the electric potentials at the surfaces of the spheres equal to each other:
Let’s assume no charges are inside the cavity
Capacitors are devices that store electric charge The capacitor is the first example of a circuit element •
A circuit generally consists of a number of electrical components (called circuit elements) connected together by conducting wires forming one or more closed loops
Substitute for the ratio of charges from Equation (1):
12
20.7 Capacitance •
A capacitor consists of two conductors of any shape •
•
•
A parallel-plate capacitor consists of two parallel plates of equal area A separated by a distance d •
•
If the capacitor is charged, one plate has charge Q and the other, charge 2Q The magnitude of the charge per unit area on either plate is = Q/A
We say that the capacitor stores charge
The capacitance, C, of a capacitor is defined as the ratio of the magnitude of the charge on either conductor to the potential difference between the conductors
• • •
The capacitance of a device depends on the geometric arrangement of the conductors •
•
20.7 Capacitance •
Capacitance is always positive The SI unit of capacitance is a farad (F) Typically you will see microfarads (F) and picofarads (pF)
20.7 Capacitance •
•
Connect two uncharged conductors to the terminals of a battery, then when disconnected, the charges remain on the conductors The conductors have charges of equal magnitude and opposite sign, and a potential difference of V between them
20.7 Capacitance •
20.7 Capacitance
If the plates are very close together (compared with their length and width), we adopt a simplification model: the electric field is uniform between the plates and zero elsewhere •
The magnitude of the field between the plates is:
•
The potential difference is:
20.7 Capacitance •
We find the capacitance:
The capacitance of a spherical conductor of radius R and charge Q is
The capacitance of an isolated charged sphere is proportional to the sphere’s radius and is independent of both the charge and the potential difference
•
The capacitance of a parallel-plate capacitor is proportional to the area of its plates and inversely proportional to the plate separation
13
20.7 Capacitance •
The field is uniform in the central region between the plates, but is nonuniform at the edges of the plates
20.7 Capacitance •
•
•
As long as the separation between the plates is small compared with the dimensions of the plates, the edge effects can be ignored
20.7 Capacitance •
Consider the circuit to be a system •
The electric potential energy is related to the separation of the positive and negative charges on the plates A capacitor can be described as a device that stores energy as well as charge
Before the switch is closed, the energy is stored as chemical energy in the battery
20.7 Capacitance •
Consider the plasma membrane for a neuron •
•
This membrane contains a number of structures, including ion channels and ion pumps, which control concentrations of various ions on either side of the membrane •
20.7 Capacitance •
When the switch is closed, the energy is transformed from chemical to electric potential energy
The plasma membrane is a lipid bilayer containing a variety of types of molecules
These ions include potassium, chlorine, calcium, and sodium
20.7 Capacitance •
There is an effective sheet of negative charge on the intracellular side of the membrane and a sheet of positive charge on the extracellular side •
•
This results in a voltage of about 70 to 80 mV across the membrane
The sheets of charge act as parallel plates, so that the membrane can be modeled as a parallel-plate capacitor •
The capacitance of the plasma membrane is about 2 F for each cm2 of membrane area
14
20.7 Capacitance •
When a neuron is carrying a signal, an event called an action potential occurs •
•
If the voltage across the membrane capacitor falls in magnitude to a threshold value of about 50 mV, the ion channels open, allowing a flow of sodium ions into the cell
This flow reduces the voltage even further, allowing more sodium ions to enter the cell, thereby reversing the polarity of the voltage across the capacitor in a time interval measured in milliseconds •
The voltage-gated ion channels then close and other channels open, allowing movement of ions until the neuron returns to its resting state
20.7 Capacitance •
This process can disturb neighboring regions of the plasma membrane so that the action potential is propagated along the neuron •
•
•
A cylindrical capacitor consists of a cylindrical conductor
•
Cylinder is coaxial with a larger cylindrical shell of radius b and charge 2Q
Special structures in the cell membrane, called voltage-gated ion channels, are normally closed
20.7 Capacitance •
20.7 Capacitance
The capacitance of the plasma membrane combines with another electrical characteristic of the membrane to provide an electrical model for the conduction of a signal along the neuron We will see how this works in the next chapter
•
•
Radius is a and charge is Q
Length is L
20.7 Capacitance •
Assume that L is large compared with a and b: •
•
adopt a simplification model in which we ignore end effects
The field is perpendicular to the axis of the cylinders and is confined to the region between them
20.7 Capacitance •
Calculate the potential difference between the two cylinders:
•
Using Gauss’s law for the inner cylinder, we know E = 2ke/r Then
•
We find
•
15
20.7 Capacitance •
The capacitance is proportional to the length of the cylinders •
•
20.8 Combinations of Capacitors •
It also depends on the radii of the two cylindrical conductors
Shown is a circuit diagram for this combination of capacitors •
The capacitance per unit length of a coaxial cable is •
20.8 Combinations of Capacitors •
A circuit diagram is a simplified representation of an actual circuit •
•
•
•
20.8 Combinations of Capacitors •
Circuit symbols are used to represent the various elements Lines are used to represent wires The battery’s positive terminal is at higher potential and is indicated by the longer line
20.8 Combinations of Capacitors The two capacitors connected as shown are known as a parallel combination of capacitors
The left plates of the capacitors are connected to the positive terminal of the battery by a conducting wire and are therefore both at the same electric potential as the positive terminal Likewise, the right plates are connected to the negative terminal and so are both at the same potential as the negative terminal
Therefore, the individual potential differences across capacitors connected in parallel are the same and are equal to the potential difference applied across the combination •
That is,
20.8 Combinations of Capacitors •
After the battery is attached to the circuit, the capacitors quickly reach their maximum charge •
•
Let’s call the maximum charges on the two capacitors Q1 and Q2 The total charge Q tot stored by the two capacitors is the sum of the charges on the individual capacitors:
16
20.8 Combinations of Capacitors •
You can replace the two capacitors with an equivalent capacitor with capacitance Ceq: •
•
20.8 Combinations of Capacitors •
The equivalent capacitor must store charge Qtot when connected to the battery The voltage across the equivalent capacitor is V, so:
When the battery is connected, electrons are transferred out of the left plate of C1 and into the right plate of C2 •
and
20.8 Combinations of Capacitors •
The equivalent capacitance for capacitors connected in parallel is:
20.8 Combinations of Capacitors •
The negative charge leaving the left plate of C2 causes negative charges to accumulate on the right plate of C1 •
•
The equivalent capacitance of a parallel combination of capacitors is
•
(1) the algebraic sum of the individual capacitances (2) greater than any of the individual capacitances
20.8 Combinations of Capacitors •
The two capacitors shown and the equivalent circuit diagram are a series combination or capacitors
As this negative charge accumulates on the right plate of C2, an equivalent amount of negative charge is forced off the left plate of C2, and this left plate therefore has an excess positive charge
As a result, both right plates end up with a charge 2Q, and both left plates end up with a charge 1Q
Therefore, the charges on capacitors connected in series are the same:
20.8 Combinations of Capacitors •
•
The total voltage Vtot across the combination is split between the two capacitors:
In general, the total potential difference across any number of capacitors connected in series is the sum of the potential differences across the individual capacitors
17
20.8 Combinations of Capacitors •
Applying the definition of capacitance to the equivalent circuit shown:
•
Substituting for voltages:
•
Canceling charges:
20.8 Combinations of Capacitors •
For capacitors connected in series, the equivalent capacitance is:
•
This expression shows that:
Example 20.8 Equivalent Capacitance •
The 1.0-F and 3.0-F capacitors (upper redbrown circle) are in parallel. Find the equivalent capacitance:
Example 20.8 Equivalent Capacitance •
The 2.0-F and 6.0-F capacitors (lower redbrown circle) are also in parallel:
(1) the inverse of the equivalent capacitance is the algebraic sum of the inverses of the individual capacitances (2) the equivalent capacitance of a series combination is always less than any individual capacitance in the combination
Example 20.8 Equivalent Capacitance Find the equivalent capacitance between a and b for the combination of capacitors shown. All capacitances are in microfarads.
Example 20.8 Equivalent Capacitance •
The two 4.0-F capacitors (upper green circle) are in series. Find the equivalent capacitance:
18
Example 20.8 Equivalent Capacitance •
The two 8.0-F capacitors (lower green circle) are also in series. Find the equivalent capacitance:
20.9 Energy Stored in a Capacitor •
The work done in charging the capacitor appears as electric potential energy U:
• •
•
Example 20.8 Equivalent Capacitance •
The 2.0-F and 4.0-F capacitors are in parallel:
20.9 Energy Stored in a Capacitor •
The energy stored in a capacitor can be modeled as being stored in the electric field between the plates of the capacitor
•
The energy per unit volume, called the energy density, is •
20.9 Energy Stored in a Capacitor •
Assume the capacitor is being charged and, at some point, has a charge q on it •
•
The work needed to transfer a charge from one plate to the other is
This applies to a capacitor of any geometry The energy stored increases as the charge increases and as the potential difference increases In practice, there is a maximum voltage before discharge occurs between the plates
The energy density in any electric field is proportional to the square of the magnitude of the electric field at a given point
Example 20.9 Rewiring Two Charged Capacitors Two capacitors C1 and C2 (where C1 > C2) are charged to the same initial potential difference Vi. The charged capacitors are removed from the battery, and their plates are connected with opposite polarity. The switches S1 and S2 are then closed.
The total work required is
19
Example 20.9 Rewiring Two Charged Capacitors
Example 20.9 Rewiring Two Charged Capacitors
(A) Find the final potential difference Vf between a and b after the switches are closed. •
Write an expression for the total charge on the left-hand plates of the system before the switches are closed:
Example 20.9 Rewiring Two Charged Capacitors •
•
Write an expression for the total charge on the left-hand plates of the system after the switches are closed:
•
•
•
A dielectric is an insulating material that, when placed between the plates of a capacitor, increases the capacitance
Solve for Vf:
(B) Find the total energy stored in the capacitors before and after the switches are closed and determine the ratio of the final energy to the initial energy.
•
Dielectrics include rubber, plastic, or waxed paper
Because the system is isolated, the initial and final charges on the system must be the same:
Example 20.9 Rewiring Two Charged Capacitors
•
Divide Equation (5) by Equation (4) to obtain the ratio of the energies stored in the system:
20.10 Capacitors with Dielectrics
•
•
Rewrite this expression in terms of Vi:
Find an expression for the total energy stored in the capacitors before the switches are closed:
20.10 Capacitors with Dielectrics •
When a dielectric material is inserted between the plates of a capacitor, the capacitance increases •
If the dielectric completely fills the space between the plates, the capacitance increases by the dimensionless factor , called the dielectric constant of the material
Write an expression for the total energy stored in the capacitors after the switches are closed:
20
20.10 Capacitors with Dielectrics •
•
Consider a parallel-plate capacitor of charge Q0 and capacitance C0 in the absence of a dielectric The potential difference across the capacitor as measured by a voltmeter is V0 = Q0/C0 •
A dielectric is inserted between the plates •
For a parallel-plate capacitor:
•
In theory, d could be made very small to create a very large capacitance In practice, there is a limit to d
•
•
•
Because the charge Q0 on the capacitor does not change, the capacitance must change to the value:
•
A dielectric provides the following advantages: • • •
An increase in capacitance An increase in maximum operating voltage Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C
20.10 Capacitors with Dielectrics •
The molecules that make up the dielectric are modeled as dipoles •
•
d is limited by the electric discharge that could occur though the dielectric medium separating the plates For a given d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the dielectric strength of the material
20.10 Capacitors with Dielectrics
The voltmeter reading decreases by a factor of to the value V, where
20.10 Capacitors with Dielectrics •
•
Notice that the capacitor circuit is open; that is, the plates of the capacitor are not connected to a battery and charge cannot flow through an ideal voltmeter
20.10 Capacitors with Dielectrics •
20.10 Capacitors with Dielectrics
The molecules are randomly oriented in the absence of an electric field
The capacitance increases by the factor when the dielectric completely fills the region between the plates
21
20.10 Capacitors with Dielectrics •
An external electric field is applied •
•
The charged edges of the dielectric act as a second pair of plates producing an induced electric field in the direction opposite the original electric field
20.10 Capacitors with Dielectrics •
Metallic foil may be interlaced with thin sheets of paper or Mylar •
•
This produces a torque on the molecules The molecules partially align with the electric field
20.10 Capacitors with Dielectrics •
20.10 Capacitors with Dielectrics
The layers are rolled into a cylinder to form a small package for the capacitor
High-voltage capacitors commonly consist of a number of interwoven metallic plates are immersed in silicon oil
20.10 Capacitors with Dielectrics •
An electrolytic capacitor can store large amounts of charge at relatively low voltages •
•
It consists of a metal foil in contact with an electrolyte When a voltage is applied between foil and electrolyte, a thin layer of petal oxide is formed, which serves as the dielectric
Example 20.10 Energy Stored Before and After A parallel-plate capacitor is charged with a battery to a charge Q0. The battery is then removed, and a slab of material that has a dielectric constant is inserted between the plates. Identify the system as the capacitor and the dielectric. Find the energy stored in the system before and after the dielectric is inserted.
22
Example 20.10 Energy Stored Before and After •
•
•
Find the energy stored in the absence of the dielectric:
Find the energy stored in the capacitor after the dielectric is inserted between the plates:
20.11 Context Connection: The Atmosphere as a Capacitor •
The capacitance is:
•
Substituting numerical values:
Replace the capacitance C:
20.11 Context Connection: The Atmosphere as a Capacitor •
•
•
A negative charge occurs on the Earth’s surface Positive charges are distributed through the atmosphere This separation of charge can be modeled as a capacitor
20.11 Context Connection: The Atmosphere as a Capacitor •
The charge distribution on the surface is assumed to be spherically symmetric, so:
•
The potential difference is
•
h is about 5 km
23
