Transcript CHAPTER 25 : ELECTRIC POTENTIAL 25.1) POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL The test charge is moved in the field by some external.

CHAPTER 25 : ELECTRIC POTENTIAL
25.1) POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL
The test charge is moved in the field by some external agent
The work done by the field on the charge = the negative of the work done by
the external agent causing the displacement.
• For an infinitesimal displacement ds, the work done by the electric field on the
charge is :




F  ds  q o E  ds
• As this amount of work is done by the field, the potential energy of the chargefield system is decreased by an amount :
 
dU  qo E  ds
• For a finite displacement of the charge from a point A to a point B, the change in
potential energy of the system ΔU = UB – UA is :


B
U  qo A E  ds
(25.1)
• Line integral does not depend on the path taken from A to B
• The electric potential :
Scalar
quantitiy
U
V
qo
Potential energy
(25.2)
charge
• The potential difference V = VB – VA between any two points A and B in an
electric field is defined as the change in potential energy of the system divided
by the test charge qo :
 
U
B
V 
  A E  d s
qo
(25.3)

• Potential difference
difference in potential energy.
• Potential difference is proportional to the change in potential energy, they are
related by :
U  q o V
Electric potential is a scalar characteristic of an electric field, independent
of the charges that may be placed in the field.
• Potential energy  the charge-field system.
• Interest  knowing the electric potential at the location of a charge
 and the potential energy resulting from the interaction of the charge
with the field.
• The change in potential energy of a charge is the negative of the work done by
the electric field on the charge (Equation (25.1))
• The potential difference V between points A and B equals the work per unit
charge that an external agent must perform to move a test charge from A to B –
without changing the kinetic energy of the test charge.
• The electric potential at an arbitrary point in an electric field equals the work
required per unit charge to bring a positive test charge from infinity to that
point.
• The electric potential at any point P is :
 
Vp    E  d s
P

(25.4)
Definition of Volt
• Electric potential is a measure of potential energy per unit charge, the SI unit of
both electric potential and potential difference is joules per coulomb, which is
defined as a volt (V) :
J
1V  1
C
• That is – 1 J of work must be done to move a 1-C charge through a potential
difference of 1V.
• The SI unit of electric field (N/C) can also be expressed in volts per meter :
N
V
1 1
C
m
• Electron volt (eV) = the energy an electron (or proton) gains or loses by moving
through a potential difference of 1V.
• 1V = 1 J/C & the fundamental charge is approximately 1.601019 C  the
electron volt is related to the joule as :
1eV  1.601019 C  V  1.601019 J
(25.5)
25.2) POTENTIAL DIFFERENCES IN A UNIFORM ELECTRIC FIELD
A
A
d
d
q
m
B
B

E

g
(a)
FIGURE (25.1)
(b)
• Calculate the
potential difference
between two points A
and B separated by a
distance d, where d is
measured parallel to
the field lines.
• Equation (25.3) gives :
 
B
B
o
VB  VA  V   E  ds  A E cos0 ds  A Eds
B
A
• E is constant :
V  EA ds  Ed
B
(25.6)
• Electric field lines always point in the direction of decreasing electric potential
(Figure 25.1a).
• Suppose that a test charge qo moves from A to B.
• The change in its potential energy from equation (25.3) and (25.6) is :
U  qo V  qo Ed
(25.7)
If qo is positive, U is negative
If qo is negative, U is positive
• Conclusion – a positive charge loses
electric potential energy when it
moves in the direction of the electric
field.
• Situation is reversed.
• If a positive test charge is released
from rest in this electric field  it
experiences an electric force qoE in
the direction of E (downward in Fig
25.1a)
• It accelerates downward, gaining
kinetic energy.
• As the charged particle gains kinetic
energy, it loses an equal amount of
potential energy.
• A negative charge gains electric
potential energy when it moves in the
direction of the electric field.
• If a negative charge is released from
rest in the field E, it accelerates in a
direction opposite the direction of the
field.
B

E
s
A
C
• A charge particle that is free to move
between any two points in a uniform
electric field directed along the x-axis
(Figure (25.2)).
FIGURE (25.2)
• Equation (25.3) gives :
 
 B 
 
V   E  ds  E  A ds  E  s
B
A
(25.8)
• The change in potential energy of the charge is :
 
U  qo V  qo E  s
(25.9)
• Conclusion – From equation (25.8), all points in a plane perpendicular
to a uniform electric field are at the same electric potential.
An equipotential surface
• Figure (25.2)
• The potential difference VB – VA is equal to the potential difference VC – VA.
• VB = VC
The name equipotential surface is given to any surface consisting of a
continuous distribution of points having the same electric potential.
Note :
• U = qo V  no work is done in moving a test charge between any two
points on an equipotential surface.
• The equipotential surfaces of a uniform electric field consist of a family of
planes that are all perpendicular to the field.
Example (25.1) : The Electric Field Between Two Parallel Plates of Opposite
Charge
A battery produces a specified potential difference between conductors attached
to the battery terminals. A 12-V battery is connected between two parallel plates,
as shown in Figure (25.4). The separation between the plates is d = 0.30 cm, and
we assume the electric field between the plates to be uniform. (This assumption is
reasonable if the plate separation is small relative to the plate dimensions and if
we do not consider points near the plate edges).
Find the magnitude of the electric field between the plates.
A
B
+ 12 V -
d
FIGURE (25.4)
Example (25.2) : Motion of a Proton in a Uniform Electric Field
A proton is released from rest in a uniform electric field that has a magnitude
of 8.0 x 104 V/m and is directed along the positive x-axis (Figure (25.5)). The
proton undergoes a displacement of 0.50 m in the direction of E.
(a) Find the change in electric potential between points A and B.
(b) Find the change in potential energy of the proton for this displacement.
FIGURE (25.5)
25.3) ELECTRIC POTENTIAL AND POTENTIAL ENERGY DUE TO
POINT CHARGES
FIGURE (25.6)
• An isolated positive point charge q.
• A charge produces an electric field that is directed radially outward from the
charge.
• The electric potential at a point located a distance r from the charge,
 
VB  VA   E  ds
B
A
• The electric field due to the point charge is
 k eq
E  2 rˆ
r
(Eq. (23.4))
 
• The quantity E  d s can be expressed as :
 

q
E  d s  k e 2 rˆ  d s
r

• rˆ  1 , therefore rˆ  d s  ds cos 

•  = the angle between rˆ and d s 

• ds cos  is the projection of d s on r - ds cos  = dr.

• That is, any displacement d s along
the path from point A to point B produces a

change dr in the magnitude of r , the radial distance to the charge creating the
field.
• Substitutions :
   k eq 
E  d s   2 dr
 r 
• The potential difference :
rB
dr k e q 
VB  VA    E r dr  k e q  2 
r
r  rA
rB
rA
1 1
VB  VA  k e q   
 rB rA 
(25.10)
Notes :
•The integral of
is independent of the path between points A and B.
 
s potential difference between any two points A and B in a
• Equation (25.10)E– dthe
field created by a point charge depends only on the radial coordinates rA and rB.
•As reference, we choose the electric potential to be zero at rA = .
•The electric potential created by a point charge at any distance r from the charge
is :
q
V  ke
r
(25.11)
Electric potential due to several point charges
• The total electric potential at some point P due to several point charges is the sum
of the potentials due to the individual charges.
• For a group of point charges, the total electric potential at P :
qi
V  ke 
i ri
(25.12)
Electric potential energy due to two charges
• The potential energy of a system of two charged particles.
• If V1 is the electric potential at a point P due to charge q1 – then the work an
external agent must do to bring a second charge q2 from infinity to P without
acceleration is :
q 2 V1
• This work equals the potential energy U of the two-particle system when the
particles are separated by a distance r12 (Figure (25.9)).
r12
q2
If more than two charged particles
in the system
FIGURE (25.9)
q1
•The potential energy is :
q1q 2
U  ke
r12
(25.13)
Note
i)
ii)
If the charges are of the same sign, U
is positive (means that positive work
must be done by an external agent on
the system to bring the two charges
near one another, because like charges
repel).
If the charges are of opposite sign, U is
negative (means that negative work
must be done against the attractive
force between the unlike charges for
them to be brought near each other.
• The total potential energy is
obtained by calculating U for every
pair of charges and summing the
terms algebraically.
• Example (Figure (25.10)), the total
potential energy of the system of
three charges is :
 q1q 2 q1q 3 q 2q 3 

U  k e 


r13
r23 
 r12
(25.14)
q2
r12
r23
q1
FIGURE (25.10)
r13
q3
Interpretation :
• Imagine that q1 is fixed at the position shown in Figure (25.10), but that q2 and q3
are at infinity.
• The work an external agent must do to bring q2 from infinity to its position near
q1 is keq1q2 / r12 , which is the first term in Equation (25.14).
• The last two terms represent the work required to bring q3 from infinity to its
position near q1 and q2
Example (25.3) : 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, as shown in Figure (25.11a).
(a) Find the total electric potential due to these charges at the point P, whose
coordinates are (4.00,0)m.
(b) Find the change in potential energy of a 3.00 C charge as it moves from
infinity to point P (Figure (25.11b))
y
y
-6.00 C
-6.00 C
FIGURE (25.11)
(a)
(b)
3.00m
3.00m
3.00 C
P
x
x
4.00m
4.00m
2.00 C
2.00 C
25.4) OBTAINING THE VALUE OF THE ELECTRIC FIELD FROM THE
ELECTRIC POTENTIAL
• Calculate the value of the electric field if the electric potential is known.
• From Equation (25.3), the potential difference dV between two points a distance
ds apart is :
 
dV  E  ds
25.15
• If the electric field has only one component Ex, then
 
E  ds  Ex dx
• Equation (25.15) becomes :
dV  E x dx
or
dV
Ex  
dx
(25.16)
• The magnitude of the electric field in the direction of some coordinate = the
negative of the derivative of the electric potential with respect to that coordinate.
• A uniform electric field produced by an infinite sheet of charge.
• Equipotential surfaces are perpendicular to the field (Fig. 25.12a).

E
Equipotential surfaces
Electric field lines
FIGURE (25.12a)
FIGURE (25.12b)
• Figure (25.12b) – a point charge
• A small positive charge placed at rest on an electric field line begins to move
along the direction of E because that is the direction of the force exerted on the
charge by the charge distribution creating the electric field.
• Figures (25.12a) and (25.12b) – a charge placed at rest in the field will move in a
straight line because its acceleration vector is always parallel to its velocity
vector.
• The magnitude of v increases, but its direction does not change.
FIGURE (25.12c)
• Figure (25.12c) – an electric dipole.
• A positive charge placed at some point near the dipole first moves in a
direction parallel to E at that point.
• Because the direction of the electric field is different at different locations,
however, the force acting on the charge changes direction, and a is no longer
parallel to v.
• This causes the moving charge to change direction and speed, but it does not
necessarily follow the electric field lines.
• If the charge distribution creating an electric field has spherical symmetry such
that the volume charge density depends only on the radial distance r, then the
electric field is radial.
 
E  ds  E r dr
• Therefore
thus,
dV
Er  
dr
dV  E r dr
(25.17)
• Note that the potential changes only in the radial direction, not in any direction
perpendicular to r.
• V (like Er) is a function only of r.
• This is consistent with the idea that equipotential surfaces are perpendicular to
field lines.
• The e1quipotential surfaces are a family of spheres concentric with the
spherically symmetric charge distribution (Figure (25.12b)
• Figure (25.12c) - the equipotential surface for an electric dipole.
• When a test charge undergoes a displacement ds along an equipotential surface,
then dV=0 because the potential is constant along an equipotential surface.

 
• Equation (25.15), dV  E  ds  0 ; E must be perpendicular to the
displacement along the equipotential surface.
• This shows that the equipotential surface must always be perpendicular to the
electric field lines.
• The electric potential is a function of all three spatial coordinates.
• If V(r) is given in terms of the cartesian coordinates, the electric field
components Ex, Ey, and Ez can readily be found from V(x,y,z) as the partial
derivatives :
V
Ex  
x
V
Ey  
y
V
Ez  
z
Example (25.4) : The Electric Potential Due to a Dipole
an electric dipole consists of two charges of wqual magnitude and opposite sign
separated by a distance 2a, as shown in Figure (25.13). The dipole is along the xaxis and is centered at the origin.
(a) Calculate the electric potential at point P
(b) Calculate V and Ex at a point far from the dipole.
(c) Calculate V and Ex if point P is located anywhere between the two charges.
y
a
-q
a
P
q
x
x
FIGURE (25.13)
25.5) ELECTRIC POTENTIAL DUE TO CONTINUOUS CHARGE
DISTRIBUTIONS
• We can calculate the electric potential due to a continuous charge distribution
in two ways :
Step 1
• If the charge distribution is known, we can start with Equation (25.11) for the
electric potential of a point charge.
• Consider the potential due to a small charge element dq, treating this element
as a point charge (Figure (25.14).
• The electric potential dV at some point P due to the charge element dq is :
dq
dV  k e
r
(25.18)
r = the distance from the charge
element to point P.
Step 1 (continue)
• The total potential at point P, we integrate Equation (25.18), to include
contributions from all elements of the charge distribution.
• Each element is - a different distance from point P and ke is constant, thus V :
dq
V  ke 
r
(25.19)
Step 2
• If the electric field is already known - caluculate the electric potential due to a
continuous charge distribution using Equation (25.3).
• If the charge distribution is highly symmetric, we first evaluate E at any point
using Gauss’s Law and then substitute the value obtained into Equation (25.3) to
determine the potential difference V between any two points.
• We then choose the electric potential V to be zero at some convenient point.
Example (25.5) : Electric Potential Due to a Uniformly Charged Ring
(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
cahrge Q.
(b) Find an expression for the magnitude of the electric field at point P.
FIGURE (25.15)
Example (25.6) : Electric Potential Due to a Uniformly Charged disk
(a) Find the electric potential and
(b) Find the magnitude of the electric field along the perpendicular central axis
of a uniformly charged disk of radius a and surface charge density .
FIGURE (25.16)
Example (25.7) : Electric Potential Due to a Finite Line of Charge.
A rod oflength  located along the x-axis has a total charge Q and a uniform linear
charge density  = Q /  . Find the electric potential at a point P located on the y
axis a distance a from the origin (Figure (25.17).
y
P
FIGURE (25.17)
r
a
0
dq
x
x
dx

Example (25.8) : Electric Potential Due to a Uniformly Charged Sphere
An insulating solid sphere of radius R has a uniform positive bolume chyarge
density and total charge Q.
(a) Find the electric potential at a point outside the sphere, that is, for r > R. Take
the potentail to be zero at r = .
(b) Find the potential at a point inside the sphere, that is, for r < R.
R
D
C
Q
r
FIGURE (25.18)
B
25.6) ELECTRIC POTENTIAL DUE TO A CHARGED CONDUCTOR
•
In section (24.4) – when a solid conductor in equilibrium carries a net charge :
i)
The charge resides on the outer surface of the conductor.
ii)
The electric field just outside the conductor is perpendicular to the surface, and
iii) The electric field inside the conductor is zero
•
Every point on the surface of a charged conductor in equilibrium is at the same
electric potential.
•
Two points A and B on the surface of a charged conductor, as shown in Figure
(25.20).
•
•

Along a surface path connecting
 these points, E is always perpendicular to the
displacement ds; therefore E  ds  0 .
Using this result and Equation (25.3), the potential difference between A and B
is necessarily zero :
 
VB  VA   E  ds  0
B
A
•
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. Furthermore, because the electric field is zero inside the
conductor, we conclude from the relationship Er = -dV/dr that the electric potential
is constant everywhere inside the conductor and equal to its value at the surface.
• No work is required to move a test charge from the interior of a charged conductor
to its surface.
(a)
• A solid metal conducting sphere of radius R and total
positive charge Q, as shown in Figure (25.21a)
+ +
+ R
+
+
+
+ +
• The electric field outside the sphere is keQ/r2 and
points radially outward.
k eQ
r
V
(b)
k eQ
R
r
E
k eQ
r2
(c)
R
FIGURE (25.21)
r
• From example 25.8 – the electric potential at the
interior and surface of the sphere must be keQ/R
relative to infinity.
• The potential outside the spher is keQ/r.
• Figure (25.21b) – a plot of the electric potential as a
function of r,
• Figure (25.21c) – shows how the electrric field varies
with r.
• When a net charge is placed on a spherical conductor, the surface charge
density is uniform (Figure (25.21a)
• If the conductor is nonspherical, Figure (25.20) – the surface charge density is
high where the radius of curvature is small and the surface is convex, and it is
low where the radius of curvature is small and the surface is concave.
• Because the electric field just outside the conductor is propotional to the
surface charge density, - the electric field is large near convex points having
small radii of curvature and reaches very high values at sharp points.
Example (25.9) : Two Connected Charged
Spheres
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, as shown in Figure (25.23).
The charges on the spheres in equilibrium are q1
and q2, respectively, and they are uniformly
charged. Find the ratio of the magnitudes of the
electric fields at the surfaces of the spheres.
q1
r1
FIGURE (25.23)
q2
r2
Charge Distribution
Uniformly charged ring
of radius a
Uniformly charged disk
of radius a
Uniformly charged,
insulating solid sphere of
radius R and total charge
Q
Isolated conducting
sphere of radius R and
total charge Q
Electric Potential
V  ke
Location
Along perpendicular central
axis of ring, distance x
from ring center.
Q
x2  a2
Along perpendicular central
1/ 2
V  2k e  x 2  a 2   x axis of disk, distance x
from disk center.
Q
V  ke
rR
r




k eQ 
r2 
 3  2 
V
2R  R 
r<R
V  ke
Q
r
r >R
V  ke
Q
R
rR