1.44×10−18 1.44×10−18 0 Every time a charged particle moves along a line of constant potential, its potential energy remains constant and the electric field does no work on the particle.
Equipotential Surfaces Surfaces in a Capacitor Description: Find the work done by a charge moving along an equipotential surface, and use that idea to find the distance between the charged plates in a parallel-plate capacitor such that there is a given potential with a given electric field between the plates. Part A Is the electric potential potential energy of a particle with with charge Hint A.1
the same at all points points on an equipotential surface? surface?
Formula for electric potential energy
For a particle with charge
at potential
ANSWER:
Yes
, the electric potential energy is
.
No For a particle with charge
on an equipotential surface at potential
, the electric potential energy has a constant value
.
Part B What is the work work required to move a charge around on an equipotential equipotential surface at at potential Hint B.1
with constant constant speed?
A formula for work
The total work done on an object is equal to the change in its energy (potential + kinetic). ANSWER: Work =
Since the speed of the charge is constant as it moves along the equipotential surface, and the electric potential energy is constant on that surface, there is no change in the total energy of the charge. This also means that no work is done by t he charge as it moves along the equipotential surface. Part C What is the work work done by the electric electric field on a charge as as it moves along an equipotential surface at potential potential Hint C.1
?
Work done by an electric field on a charged particle
The force due to an electric field is a conservative force. As such, the work done by such a force is equal to the change in the potential energy of the particle it is acting on. ANSWER: Work done by the electric field =
Just as in Part B, since there is no change in the electric potential energy, no work is done by the electric field as the charge moves along the equipotential surface. Part D The work
done by the uniform electric field
in displacing a particle with charge
along the path
is given by
,
where
is the angle between
and
Express your answer in radians. ANSWER:
. Since in general,
is not equal to zero, for points on an equipotential surface, what must
be for
to equal 0?
=
You have shown that equipotential surfaces are always perpendicular to the electric field at their surface.
Now assume that a parallel-plate parallel-plate capacitor capacitor is attached across across the terminals of a battery as shown shown in the figure. The electric electric field
in the region between the plates points in the negative negative z direction, from higher to lower voltage.
Part E Find the electric potential Hint E.1 ield
at a point
inside the capacitor if the origin of the coordinate system
is at potential .
The relation between electric potential and the electric field
in a region of space is
,
where the line integral may be taken along any path . Hint E.2
Expressing an infinitesimal length element
In general, a small length vector along the path of choice can be written as .
Substitute this expression expression into the integral for
Analysis of the equation
Hint E.3 Recall that if
.
and
are perpendicular (where
is one of the Cartesian coordinate axes), then
Express your answer in terms of some or all of the variables ANSWER:
,
, since
. According to the setup of this part, only one of the directions ( , , ) will not be perpendicular to the electric field as defined.
, , and .
=
Therefore, the equation equation of an equipotential equipotential surface at a potenial
is given given by
.
This is the equation of a plane that is parallel parallel to the plates of the capacitor capacitor and perpendicular to the electric electric field. In particular, the the lower plate, which which is at zero potential, potential, corresponds to the surface surface
.
Part F What is the distance Hint F.1
between two surfaces separated by a potential difference
?
How to approach the problem
Use the equation you found in Part E to find equations that represent the potentials Express your answer in terms of
and
and
of the planes located at
and
. Use these expressions to find an equation for
.
.
ANSWER: =
The Fate of an Electron in a Uniform Electric Field Description: Review concept of electric potential and electric potential energy for an electron between two capacitor plates. Compute the speed of the electron when it reaches the upper plate after being released from rest in the middle of the capacitor. Compute how this speed changes when the mass or charge of the "electron" is changed. In this problem we we will study the behavior of an electron electron in a uniform electric field. Consider a uniform electric electric field
First, let us review the the relationship between between an electric field and its associated associated electric potential potential
(magnitude
) as shown in the the figure within a parallel plate capacitor in vacuum. vacuum.
. For now, ignore the electron electron located between between the plates.
Part A Calculate the electric electric potential Hint A.1
inside the capacitor capacitor as a function of height height
Relationship of field and potential
The general relation between the electric potential in an electric field is
Hint A.2
. Take Take the potential at the bottom plate to be zero.
Limits of integration
. Because electric fields are gradient (or conservative) fields, we can write this relation in an integral form:
.
Integrate from the bottom plate to arbitrary height Express
in terms of
and
. You know the value of the the potential at the the bottom plate: It is zero. Be careful with signs! You are integrating against the direction of
.
ANSWER:
Now an electron of mass
, where
=
and charge
(where
is a positive quantity) is placed within within the electric field (see the figure) at height
.
Part B Calculate the electon's potential energy Hint B.1
, neglecting gravitational potential energy.
Definition of electric potential energy
The definition of the electric potential is that it is equal to the potential energy per unit charge. Therefore, Express your answer in terms of
,
ANSWER:
, and
for a charge
.
.
=
Part C The electron, electron, having been held at height Hint C.1
, is now released from rest. rest. Calculate Calculate its speed speed
(i.e.,
) when when it reaches the top plate. plate.
How to approach the problem
Although this problem can be done in several different ways, the easiest (that is, with the least amount of calculation) is to use which of the following principles? ANSWER:
conservation of energy with gravitational potential energy and kinetic energy conservation of energy with electrostatic potential energy and kinetic energy conservation of energy with gravitational and electrostatic potential energy Newt Newton on's 's seco second nd law law ( ) fol follo lowe wed d by kine kinema mati tics cs
The only two forms of energy that we are considering in this problem (since we have excluded gravity) are the electric potential energy and the kinetic energy. Find the initial energy
Hint C.2
Find an expression for the initial energy Express
in terms of
,
,
(kinetic plus potential) of the electron at height
, and
ANSWER:
.
.
=
Find the final energy
Hint C.3
Assume that that the speed of the electron electron when itit reaches reaches the upper plate plate is . Find Find an expression for the final energy energy Hint C.3.1
Kinetic energy of a point particle
The kinetic energy of a point particle of mass
Express
in terms of
,
,
,
ANSWER: =
,
moving with velocity
, and .
is
.
(kinetic plus potential) potential) of the electron electron when when it reaches the upper plate (at position position
).
, which affects the sign.
Part A What property of objects is best measured by their capacitance? ANSWER:
ability to conduct electric current ability to distort an external electrostatic field ability to store charge ability to store electrostatic energy
Capacitance is a measure measure of the ability of a system of two conductors to store electric electric charge and energy. Capacitance is defined as as
. This ratio ratio remains constant constant as long as the system retains its geometry and the amount of dielectric dielectric does not change. Capacitors
are special devices designed to combine a large capacitance capacitance with a small size. However, However, any pair of conductors separated by a dielectric (or vacuum) has some capacitance. capacitance. Even an isolated electrode has a small capacitance. capacitance. That is, if a charge potential
with respect to ground will change, and the ratio
is its capacitance capacitance
.
Part B Assume that charge
is placed on the top plate, and
Hint B.1
is placed on the bottom plate. What is the magnitude of the electric field
between the plates?
How do you find the magnitude of the electric field?
What is the easiest way to obtain obtain ANSWER:
? Use Gauss's law and the fact that
outside the capacitor.
Use Gauss's law and the symmetry of the lower plate. Use Coulomb's law integrating over all charge on the bottom plate. Use Coulomb's law integrating over all charge on both plates.
Hint B.2
What is the electric flux integral due to the electric field?
Apply Gauss's Gauss's law to a small box whose whose top side side is is just above the lower plate and whose whose bottom is just just below below it, it, where Express this integral in terms of the area ANSWER:
of the top side of the box and the magnitude
. Start Start by finding finding the electric flux integral integral
of the electric field between the plates.
=
Hint B.3
Find the enclosed charge
Find the amount of charge Hint B.3.1
enclosed in a small box whose top side is just above the lower plate and whose bottom is just below it, where Find the surface charge
What is , the charge charge per unit unit area on the lower lower plate? Express
in terms of any necessary constants constants and quantities given in the the introduction. introduction.
ANSWER: =
Express the enclosed charge charge in terms of the cross-sectional cross-sectional area of the box
and other quantities quantities given in the introduction. introduction.
ANSWER: =
Hint B.4
Recall Gauss's law
Gauss's law states that
Express
in terms of
.
and other quantities given in the introduction, in addition to
and any other constants needed. needed.
.
.
is placed on it, its
ANSWER: =
Part C What is the voltage
between the plates plates of the capacitor? capacitor?
Hint C.1
The electric field is the derivative of the potential
The voltage difference is the integral of the electric field from one plate to the other; in symbols, Express
.
in terms of the quantities quantities given in the introduction introduction and any required required physical constants. constants.
ANSWER: =
Part D Now find the capacitance capacitance Express
of the parallel-plate capacitor.
in terms of quantities given in the introduction introduction and constants like
.
ANSWER: =
You have derived the general formula for the capacitance capacitance of a parallel-plate capacitor with plate area
and plate separation . It is worth remembering. remembering.
Part E Consider an air-filled charged capacitor. How can its capacitance be increased? Hint E.1
What does capacitance depend on?
Capacitance depends on the inherent properties of the system of conductors, such as its geometry and the presence of dielectric, not on the charge placed on the conductors. ANSWER:
Increase the charge on the capacitor. Decrease the charge on the capacitor. Increase the spacing between the plates of the capacitor. Decrease the spacing between the plates of the capacitor. Increase the length of the wires leading to the capacitor plates.
Part F Consider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance? ANSWER:
Double the charge and double the plate area. Double the charge and double the plate separation. Halve the charge and double the plate separation. Halve the charge and double the plate area. Halve the plate separation; double the plate area. Double the plate separation; halve the plate area.
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