PHYSICS for Scientists and Engineers 讲义 ch22[1]

r 2aq ˆ ˆ +(sin θ)θ E= r 3 4πε 0 r
•
(
)
The Bottom Line
If we know the electric field E everywhere,
r r VB − VA = − ∫ E • d l
B A
allows us to calculate the potential function V everywhere (define VA = 0 above)
V(r) = q q r2 − r1 1 q − = 4 πε 0 r1 r2 4 πε 0 r1 r2
z +q aθ a -q r
≈r2-r1
r1 r2
•
Rewrite this for special case r>>a:
r2 − r1 ≈ 2 a cos θ
r1r2 ≈ r
V(r) ≡ Vr − V∞
• for a point charge, the formula is:
V(r) ≡ Vr − V∞ =
1 q 4 πε 0 r
Potential from charged spherical shell
V
• E Fields (from Gauss' Law)
• r < R: • r > R:
Potential from N charges
The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately.
r=r N r r r V(r) = − ∫ E • d l = − ∫ ∑ E n • d l r=∞ r = ∞ n =1 r = rr
E=0
Q 4πε0 R
Q 4πε0 rE=Leabharlann 1 Q 4 πε0 r 2
R R R
r
• Potentials • r > R:
r r r 1 Q V(r ) = − ∫ E • d l = − ∫ E(dr ) = 4πε0 r r =∞ ∞ r =r
• r < R:
r R r r r 1 Q V(r ) = − ∫ E • d l = − ∫ E(dr ) = − ∫ E(dr ) − ∫ E(dr ) = +0 4πε0 R r =∞ R ∞ ∞ r =r
• We can obtain the electric field E from the potential V by inverting our previous relation between E and V:
Ex = −
∂V ∂x
Ey = −
∂V ∂y
Ez = −
∂V ∂z
•
Expressed as a vector, E is the negative gradient of V
E B
•
Since there will be a force on the charge due to E, a certain amount of work WAB will have to be done to accomplish this task. We define the electric potential difference as:
2
⇒
V(r) =
1 2 aq cos θ 4 πε 0 r2
Can we use this potential somehow to calculate the E field of a dipole?
(remember how messy the direct calculation was?)
E from V?
∴
WAB
r r r r = − ∫ Felec • dl = − ∫ q0 E • dl
B B A A
A
B
⇒
B r r WAB VB − VA ≡ = − ∫ E • dl q0 A
1
Chapter 22, ACT 1
• 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? (VAB ≡ VB - VA ) E -1µC
Electric Potential
Q 4πε0 R R R V Q 4πε0 r r
C B r
B
R
q
r
A equipotentials
A
path independence
Overview
• Introduce Concept of Electric Potential
– Is it well-defined? i.e. is Electric Potential a property of the space as is the Electric Field?
• Calculating Electric Potentials
– Charged Spherical Shell – N point charges – Electric Dipole
• Can we determine the Electric Field if we know the Electric Potential?
r r E = −∇ V
•
Cartesian coordinates:
r ∂V ∂V ∂V ∇V = x y z ˆ+ ˆ+ ˆ ∂x ∂y ∂z
•
Spherical coordinates:
r 1 ∂V ˆ 1 ∂V ˆ ∂V ∇V = θ+ φ r ˆ+ ∂r r ∂θ r sin θ ∂φ
E from V: an Example
VB − V A = E (cos θ ) r = Eh
• So here we have at least one example of a case in which the integral is the same for BOTH paths.
Electric Potential
• Define the electric potential of a point in space as the potential difference between that point and a reference point. • • a good reference point is infinity ... we typically set V∞ = 0 the electric potential is then defined as:
q1 q2 r2
r1
x r3 q3
⇒
N q 1 V ( r ) = ∑ Vn ( r ) = ∑ n 4 πε 0 n = 1 rn n =1
N
Electric Dipole
The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms.
Does it really work?
• Consider case of constant field:
– Direct: A - B
B h A
C
Θ r
dl
E
r VB − VA = − ∫ E • d l = Eh
A
Br
• Long way round: A - C - B
C r r Br r VB − V A = − ∫ E • d l − ∫ E • d l = − ∫ (− E ( dl ) cos θ ) − 0 A C A C
• This equation also serves as the definition for the potential
•The integral is the sum of the tangential (to the path) component of the electric field along a path from A to B. •The question now is: Does this integral depend upon the exact path chosen to move from A to B? •If it does, we have a lousy definition. • Hopefully, it doesn’t. • It doesn’t. But, don’t take our word, see appendix and following example.
• Consider the following electric potential:
V(x, y, z) = 3x 2 + 2xy − z 2
• What electric field does this describe? ∂V ∂V ∂V E = − = − 2 x Ex = − = −6 x − 2 y Ez = − = 2z y ∂y ∂x ∂z r ... expressing this as a vector: E = ( −6 x − 2 y ) x ˆ − 2x y ˆ+ 2z z ˆ Something for you to try: Can you use the dipole potential to obtain the dipole field? Try it in spherical coordinates ... you should get:
VB − VA ≡ WAB q0
•
Is this a good definition? • • Is VB - VA independent of q0? Is VB - VA independent of path?
Independent of Charge?
•
Fwe supply = -Felec To move a charge in an E field, we must supply a force just equal and opposite to Felec q0 that experienced by the charge due to E the E field.
Text Reference: Chapter 22
Electric Potential
• Suppose charge q0 is moved from pt A to pt B through a region of space described by electric field E.
q0 A
r VAB ≡ VB − VA = − ∫ E• dl
A
Br
Since E• dl > 0 , VAB <0 !!
r
r
Independent of Path?
Br r WAB VB − VA ≡ = −∫ E • dl q0 A
Felec
-Felec
q0 A
E B
difference VB - VA.
×
B
×
A x dl
(a) VAB < 0
(b) VAB = 0
(c) VAB > 0
• The simplest way to get the sign of the potential difference is to imagine placing a positive charge at point A and determining whether positive or negative work would be done in moving the charge to point B. • 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 can also determine the sign directly from the definition: