Transcript Electric Potential Energy

Electric Potential
Electric Potential Energy
Electric Potential Energy
Work done by Coulomb force when q1
moves from a to b:
r
dr
q1 (+) ds
FE
rb
a q1 (+)
ra
q2 (-)
b
Electric Potential Energy
r
dr
The important point is that the work
depends only on the initial and final
positions of q1.
a
ra
FE
b
ds
rab
q2 (-)
In other words, the work done by the electric force is
independent of path taken. The electric force is a
conservative force.
q1 (+)
Electric Potential Energy
A charged particle in an electric
field has electric potential
energy.
It “feels” a force (as given by
Coulomb’s law).
It gains kinetic energy and loses
potential energy if released. The
Coulomb force does positive
work, and mechanical energy is
conserved.
++++++++++++++
+
E
F
-------------------
Electric Potential
Dividing W by Q gives the potential energy per unit charge.
VAB, is known as the potential difference between points A and
B.
The electric potential V is independent of the test charge q0.
Electric Potential
++++++++++++++
If VAB is negative, there is a
loss in potential energy in
moving Q from A to B; the work
is being done by the field.
if it is positive, there is a gain
in potential energy; an external
agent performs the work
+
E
F
-------------------
Electric Potential
VAB is the potential at B with reference to A
VB and VA are the potentials (or absolute potentials) at B
and A
Electric Potential
If we choose infinity as reference the potential at infinity is
zero;the electric potential of a point charge q is
1 q
V r 
.
4 0 r
The potential at any point is the potential difference between
that point and a chosen point in which the potential is zero.
Things to remember about electric potential:
 Electric potential difference is the work per unit of charge that
must be done to move a charge from one point to another
without changing its kinetic energy.
● Sometimes it is convenient to define V to be zero at the earth
(ground).
 The terms “electric potential” and “potential” are used
interchangeably.
 The units of potential are joules/coulomb:
1 joule
1 volt =
1 coulomb
Example: a 1 C point charge is located at the origin and a -4
C point charge 4 meters along the +x axis. Calculate the
electric potential at a point P, 3 meters along the +y axis.
y
 q1 q 2 
qi
VP = k  = k  + 
i ri
 r1 r2 
-6
-6


1×10
-4×10
9
= 9×10 
+

5 
 3
P
3m
q1
4m
q2
Thanks to Dr. Waddill for the use of these examples.
x
= - 4.2×103 V
Example: how much work is required to bring a +3 C point
charge from infinity to point P?
0
y
Wexternal  q3  VP  V 
q3
P
Wexternal  3 106  4.2 103 
3m
q1
4m
q2
x
Wexternal  1.26 103 J
The work done by the external force was negative, so the work done by the electric field was
positive. The electric field “pulled” q3 in (keep in mind q2 is 4 times as big as q1).
Positive work would have to be done by an external force to remove q3 from P.
Electric Potential of a Charge Distribution
1
Collection of charges: VP 
4 0
qi
i r .
i
P is the point at which V is to be calculated, and ri is the distance of the ith
charge from P.
Charge distribution:
1
dq
V
.

40 r
Potential at point P.
dq
P
r
Electric Potential of a Charge Distribution
Example: A rod of length L located along the x-axis has a total
charge Q uniformly distributed along the rod. Find the electric
potential at a point P along the y-axis a distance d from the
origin.
y
=Q/L
P
d
dq=dx
r
dq
dx
x
L
x
dq
dx
dV  k
k
r
x2  d2
L
V   dV
0
Thanks to Dr. Waddill for this fine example.
V
L
0
dx
Q L dx
k
k 
2
2
L 0 x2  d2
x d
y
A good set of math tables will
have the integral:
P
d
r
dq
dx
x
L
x

dx
x d
2
2

 ln x  x 2  d 2
kQ  L  L2  d 2
V
ln 
L 
d





Example: Find the electric potential due to a uniformly charged
ring of radius R and total charge Q at a point P on the axis of
the ring.
dQ
r
R
P
x
Every dQ of charge on the
ring is the same distance
from the point P.
x
dq
dq
dV  k
k
r
x2  R2
V
ring
dV  k 
ring
dq
x2  R 2
dQ
r
R
P
x
x
V
V
k
x R
2
2
kQ
x2  R2

ring
dq
Example: A disc of radius R has a uniform charge per unit area
 and total charge Q. Calculate V at a point P along the central
axis of the disc at a distance x from its center.
dQ
r
P
R
x
x
The disc is made of
concentric rings. The
area of a ring at a
radius r is 2rdr, and
the charge on each ring
is (2rdr).
We *can use the equation for the potential due to a ring,
replace R by r, and integrate from r=0 to r=R.
dVring 
k2rdr
x2  r2
dQ
r
P
R
x
x
1
V   dV 
ring
40

V
x2  r2
20
R
0


20

2rdr

ring


x 2  r 2 20

Q
x R x 
20 R 2
2
2


Q

R 2
R
0
rdr
x2  r2
x2  R2  x

dQ
r
P
R
x
x
Q
V
20 R 2

x2  R 2  x

Could you use this expression for V to calculate E? Would you
get the same result as I got in Lecture 3?
MAXWELL'S EQUATION
• The line integral of E along a closed path
is zero
• This implies that no net work is done in
moving a charge along a closed path in an
electrostatic field
MAXWELL'S EQUATION
• Applying Stokes's theorem
• Thus an electrostatic field is a conservative
field
Electric Potential vs. Electric Field
• Since
we have
As a result; the electric field intensity is the gradient of V
• The negative sign shows that the direction of E is opposite to
the direction in which V increases
Electric Potential vs. Electric Field
• If the potential field V is known, the E can
be found
Example: In a region of space, the electric potential is V(x,y,z)
= Axy2 + Bx2 + Cx, where A = 50 V/m3, B = 100 V/m2, and C =
-400 V/m are constants. Find the electric field at the origin
V
E x (0, 0, 0)  
   Ay 2  2Bx  C 
 C
(0,0,0)
x (0,0,0)
V
E y (0,0,0)  
 (2Axy) (0,0,0)  0
y (0,0,0)
V
E z (0, 0, 0)  
0
z (0,0,0)
V

E(0,0,0)   400  ˆi
m
