Download Lecture 9

rb
Lecture 9: ELECTRIC POTENTIAL
Wa→b - Work by force to move the charge from r
a
a to b
q
Wa→b ΔU (U a − U b )
=
=
= Va − Vb
qo
qo
qo
E
dl
qo
Example; Calculate voltage
difference between two
positions relative to a point
charge q
Use the line integral definition
SO, the work per unit charge to move from a to b equals, V − V
a
b
r
r
rb r r rb
r r
Wa →b = ∫ F ⋅ dl = ∫ qo E ⋅ dl
Also
r
r
ra
ra
r
rb r r
General. Can
Wa →b
= Va − Vb = ∫ E ⋅ dl
find V from
r
qo
ra
E.
The potential difference, Va − Vb , is a line integral
r
q
E = K 2 rˆ
r
E
+
q
ra
dl
rb
r
rb r
r
Va − Vb = ∫ E ⋅ dl
r
ra
r
b r
rb r
r rb
Va − Vb = ∫ E • dl = ∫ E r • dr = ∫ E r dr cos0 o
a
=∫
rb
ra
ra
ra
1
1
1 1
q
q
q
(− ) | rrb =
( − )
dr =
4πε o r 2
4πε o r a 4πε o ra rb
To determine potential energy of charge q0 at point P.
U = q0 V
V =
Preflight 9:
Electric Potential
U =
qq 0 1
( )
4πε o r
A
E
Signs can be confusing. Remember:
q
1
( )
4πε o r
C
B
E
a
b
Move from a to b, in direction of E, V decreases.
Work done by electric force is positive. U decreases.
r
rb
r r
W a →b
= Va − Vb = ∫ E ⋅ dl
r
qo
ra
4) Points A, B, and C lie in a uniform electric field. What is the
potential difference between points A and B?
ΔVAB = VB - VA
a) ΔVAB > 0
b) ΔVAB = 0
c) ΔVAB < 0
5) Point C is at a higher potential than point A.
True
Preflight 9:
False
Preflight 9:
13) If you want to move in a region of electric field without
changing your electric potential energy. How would you choose
the direction ?
9) A positive charge is released from rest in a region of
electric field. The charge moves:
a) towards a region of smaller electric potential
b) along a path of constant electric potential
c) towards a region of greater electric potential
You would have to move
perpendicular to the field
if you wish to move without changing
electric potential.
1
Lecture 9, ACT 2
• A single charge ( Q = -1μC) is fixed at
the origin. Define point A at x = + 5m
and point B at x = +2m.
– What is the sign of the potential difference
Lecture 9, ACT 2
B
A
´
´
-1μC
x
• A single charge ( Q = -1μC) is fixed at
the origin. Define point A at x = + 5m
and point B at x = +2m.
– What is the sign of the potential difference
between A and B? (Δ VAB ≡ VB - VA )
(a) ΔVAB < 0
(b) Δ VAB = 0
B
A
´
´
x
-1μC
between A and B? (Δ VAB ≡ VB - VA )
(c) Δ VAB > 0
(c) Δ VAB > 0
(b) Δ VAB = 0
(a) ΔVAB < 0
•The simplest way to get the sign of the potential difference is to imagine
placing a positive charge at point A and determining which way it would move.
Remember that a positive charge will always “fall” to lower potential.
•A positive charge at A would be attracted to the -1μC charge;
therefore NEGATIVE work would be done to move the charge from A to B.
•You could also determine the sign directly from the definition:
r
r
Ar
r
ΔVAB =VB −VA = ∫ E•dl
Since E• dl > 0 , ΔVAB <0 !!
B
Δ VAB is Independent of Path
Δ VAB = VB −VA =
WAB A r r
= E • dl
q0 ∫B
q0
Does it really work?
• Consider case of constant field:
– Direct: A - B
E
A
B
B
h
A
r r A
VB −VA = ∫ E • dl = ∫ E dl = Eh
B
• The integral is the sum of the tangential (to the path) component
of the electric field along a path from A to B.
A
B
C
r
θ
E
dl
• Long way round: A - C – B
A r
r Cr r A
VB −VA = ∫ E • dl + ∫ E • dl = ∫ (E dlcosθ ) − 0
• This integral does not depend upon the exact path chosen to
move from A to B.
C
• Δ VAB is the same for any path chosen to move from A to B
B
C
V B − V A = E (cos θ ) r = Eh
(because electric forces are conservative).
• So here we have at least one example of a case in which the
integral is the same for BOTH paths.
• In fact, it works for all paths.
Preflight 9:
Lecture 9, ACT 3
A
E
B
C
7) Compare the potential differences between points A to C
and points B to C.
3
A positive charge Q is moved from A to B
along the path shown by the arrow. What
is the sign of the work done to move the
charge from A to B?
(a) WAB < 0
(b) WAB = 0
A
B
(c) WAB > 0
a) VAC > VBC
b) VAC = VBC
c) VAC < VBC
2
Electric Potential: where is it zero?
Lecture 9, ACT 3
• So far we have only considered potential differences.
A positive charge Q is moved from A to B
along the path shown by the arrow. What
is the sign of the work done to move the
charge from A to B?
3
(a) WAB < 0
•
A
B
• Define the electric potential of a point in space as the
potential difference between that point and a reference
point.
(c) WAB > 0
(b) WAB = 0
• a good reference point is infinity ... we often set V∞ = 0
A direct calculation of the work done to move a positive
charge from point A to point B is not easy.
• the electric potential is then defined as:
• Neither the magnitude nor the direction of the field is
constant along the straight line from A to B.
•
• for a point charge at origin, integrate in from infinity along some
∞r
axis, e.g., the x-axis
r
But, you DO NOT have to do the direct calculation.
•
V(r) −V(∞) = ∫ E• dl
• here “r” is distance to origin
Remember: potential difference is INDEPENDENT OF
THE PATH!!
Therefore we can take any path we wish.
Choose a path along the arc ofr a circle
centered at
r
the charge. Along this path E • dl = 0 at every point!!
•
V ( r ) ≡ Vr − V∞
r
V ( r ) ≡ Vr −V∞ =
1 q
4πε0 r
line
integral
Potential from a point charge q
Preflight 9:
Potential from N charges
+5 μC
-3 μC
A
The potential from a collection of N charges is just
the algebraic sum of the potential due to each charge
separately (this is much easier to calculate than the
net electric field, which would be a vector sum).
11) Two charges q1 = + 5 μC, q2 = -3μC are placed at equal distances
from the point A. What is the electric potential at point A?
a) VA < 0
b) VA = 0
Q1
Q2
r1p
r2p
⇒
Q3
r3p
Vat P =
N
1
n =1
4πε 0
V ( r ) = ∑ Vn ( r ) =
N
qn
n =1
n
∑r
c) VA > 0
1 Q1
1 Q2
1 Q3
+
+
4πεo r1 p 4πεo r2 p 4πεo r3 p
P
Lecture 9, ACT 4
Lecture 9, ACT 4
Which of the following charge distributions produces V(x) = 0
for all points on the x-axis? (we define V(x) ≡ 0 at x = ∞ )
Which of the following charge distributions produces V(x) = 0
for all points on the x-axis? (we define V(x) ≡ 0 at x = ∞ )
+2μC
+1μC
-2μC
-1μC
+2μC
+1μC
x
(a)
+2μC
-2μC
x
-1μC
-2μC
(b)
+2μC
+1μC
-2μC
-1μC
(c)
+1μC
+1μC
x
x
-1μC
+2μC
(a)
+2μC
-2μC
x
-1μC
-2μC
(b)
x
-1μC
(c)
+1μC
The key here is to realize that to calculate the total potential at a point, we
must only make an ALGEBRAIC sum of the individual contributions.
Therefore, to make V(x)=0 for all x, we must have the +Q and -Q
contributions cancel, which means that any point on the x-axis must be
equidistant from +2μC and -2μC and also from +1μC and -1μC.
This condition is met only in case (a)!
3
Summary
•Potential energy stored in a static charge distribution
–work we do to assemble the charges
•Electric potential energy of a charge in the presence of
a set of source charges
–potential energy of the test charge equals the potential from
the sources times the test charge: U = qV
• If we know the electric field E,
A
r r
VB − V A = ∫ E • dl
B
allows us to calculate the potential function V
everywhere (define VA = 0 above)
• Potential due to n charges:
N
V ( r ) = ∑Vn ( r ) =
n =1
1
4πε 0
N
qn
n =1
n
∑r
4