Potential difference and electric potential
Potential difference in a uniform electric field
Electric potential and electric potential energy due
to point charges
Obtaining the value of the electric field from the
electric potential
Electric potential due to continuous charge
distributions
Electric potential due to a charged conductor
Chapter 25 Electric potential
n
n
n
n
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n
†
Æ
Æ
Electric potential energy
Æ
1. The electric force Fe = q E by Coulomb’s law is conservative,
similar to the force of gravity and the elastic force exerted by a
spring.
2. The work done by the field on the charge leads to a decrease of
potential †
energy of the charge-field system, that is
Æ
dW = Fe • d s = -dU
Æ
3. For finite displacement of the charge from a point A to a point
B, the change in potential energy of the system is
BÆ
U B -U A = -q Ú E• d s
A
4. This line integral does not depend on the path from A to B.
Electric potential
1. Electric potential is equal to the electric potential energy per
unit charge U/q.
2. Electric potential has a unique value at every point in an electric
field; electric potential is a scalar characteristic of electric field.
In contrast, electric potential energy belongs to the charge-field
system.
3. Potential difference between any two points A and B in an
Æ
BÆ
electric field is given by
V
B -VA = - Ú E• d s
A
Æ
4. If one sets potential to zero at a point at infinity, the potential at
P Æ
any given point P is
VP = - Ú E• d s
•
†
5. The unit of potential is volt (V): 1V=1 J/C
A
BÆ
A
DU = qDV = -qEd
A
B
A
Æ
= - Ú E cos 0 ods = -E Ú ds = -Ed
B
DV = V -V = - Ú E• d s
B
Potential difference in a uniform
electric field
†
B
B
A
VB = VC
Æ
A
Æ
A
Æ
Æ
DU = qDV = -q E• s
Æ
= - E• Ú d s = - E• s
Æ
Æ
DV = V -V = - Ú E• d s
BÆ
Potential difference in a uniform
electric field
†
The equipotential surface is given to any surface where every point
†
has the same potential. The electric field lines are always
perpendicular to the equipotential surface.
Æ
Electric potential due to point
charges
BÆ
VB -VA = - Ú E• d s
A
Æ
Æ
qŸ Æ
q
E• d s = k
r• d s = k
dr
e
e
r2
r2
rB
dr k q rB
VB -VA = -ke q Ú = e ]
2
r rA
rA r
kq kq
= e - e
rB
rA
For rA Æ •,VA = 0
kq
kq
V = e or V = e
B
rB
r
Electric potential due to point
charges
Electric potential and potential energy
Potential for a group of point charges:
kq
V =Â e i
ri
i
Potential due to continuous charge distribution:
dq
V =k Ú
e
r
Potential energy for two point charges:
kqq
U= e 1 2
r12
Potential energy for three point charges:
kqq kq q kq q
U= e 1 2+ e 2 3+ e 3 1
r12
r23
r31
VB -VA = -Ed = -4.0 ¥10 4 V
(b) U B -U A
U B -U A = q(VB -VA )
(a) Find VB -VA
Example 25.2 motion of a proton
in a uniform electric field
Uniform electric field: 8.0¥104
V/m directed along the positive x
axis. The proton undergoes a
displacement of 0.50 m in the
direction of the field.
= -qEd = -6.4 ¥10 -15 J
(c) Find the speed of proton at B
1
U + 0 = mu 2 +U B
A
2
1
mu 2 = U A -U B = 6.4 ¥10 -15 J
2
u = 2.77 ¥10 6 m / s
= -6.29 ¥10 3 V
Total pontential energy:
kqq kq q
U= e 1 2+ e 2 3
r12
r23
kqq
+ e 3 1 = -5.48 ¥10 -2 J
r31
kq kq
VP = e 1 + e 2
r1
r2
Example 25.3 The electric
potential and potential energy
†
Æ
Relation between electric field
and electric potential
Æ
Potential difference: dV = - E• d s
If the electric field has only one component E x, then
Æ
Æ
dV
E• d s = E dx, dV = -E dx, and thus E = x
x
x
dx
In terms of the cartesian coordinates, the electric field components:
dV
dV
dV
E =- , E =- , E =x
y
z
dx
dy
dz
In terms of polar coordinates, the electric field along the radial
dV
direction: E = dr
r
Example 25.4 The electric
potential due to a dipole
kq kq
kq
kq
2k qa
V = e 1+ e 2 = e - e = e
r1
r2
x - a x + a x 2 - a2
2k qa
For x >> a, V ª e
P
x2
dV 4k qa
E == e
x
dx
x3
For P located between two charges,
kq
kq
2k qx
V= e - e = e
a - x x + a x 2 - a2
dV
x 2 + a2
E == -2k q
e
dx
(x 2 - a 2 )2
x
Example 25.5 Electric potential
due to a uniformly charged ring
dq
dq
V = ke Ú
= ke Ú
r
x 2 + a2
k
keQ
e
=
Ú dq =
x 2 + a2
x 2 + a2
dV
d
1
Ex = = -keQ (
)
dx
dx x 2 + a 2
1
= -k Q(- )(x 2 + a 2 )-3/2 (2x)
e
2
keQx
=
(x 2 + a 2 )3/2
For x = 0,
E x = 0 and V has a maximum.
Electric potential due to a
charged conductor
Electric potential is the same everywhere in a charged conductor
when it is in electrostatic equilibrium.
A
BÆ
A
Æ
V -V = - Ú E• d s = 0
B
The electric field inside
the cavity must be zero
regardless of the charge
distribution outside the
conductor.
A cavity within a conductor
†
Example 25.9 Two connected
charge spheres
Find the electric fields on
The surfaces of the spheres.
kq kq
V= e 1= e 2
r1
r2
q1 r1
=
q2 r2
q
q
E1 = ke 12 and E2 = ke 22
r1
r2
E r
1
= 2
E2 r1
This can explain why electric field
is very high at sharp points where the
radius of curvation is small.