Idea Transcript
Electric Potential Equipotentials and Energy Today: Mini-quiz + hints for HWK
U = qV
Should lightening rods have a small or large radius of curvature ? V is proportional to 1/R. If you want a high voltage to pass through the rod then use a small radius of curvature.
Air is normally an insulator, however for large E fields (E>3 x 106 V/m) it starts conducting. Empire State Building, NYC
Electrical Potential
Review:
Wa → b = work done by force in going from a to b along path.
Wa → b = ∫
b
a
r r b r r F • dl = ∫ qE • dl a
∆U = U b − U a = − Wa → b
b
F
r r = − ∫ qE • dl b
a
θ a
U = potential energy
dl
r Wa → b b r ∆U U b − U a ∆V = Vb − Va = = =− = − ∫ E • dl a q q q • Potential difference is minus the work done per unit charge by the electric field as the charge moves from a to b. • Only changes in V are important; can choose the zero at any point. Let Va = 0 at a = infinity and Vb → V, then:
r r V = − ∫ E • dl r
∞
V = electric potential allows us to calculate V everywhere if we know E
Potential from charged spherical shell • E-field (from Gauss' Law) •
r < R: Er = 0
•
r >R:
V q 4πε πε0 R
1
q Er = 4 πε 0 r 2
q 4πε πε0 r
R
r
R
• Potential
R
• r > R: r=r
V(r )= −
∫
r r r E • d l = − ∫ E r ( dr ) =
r =∞
∞
1 4πε
0
q Q r
• r < R: r aR r r r 1 V ( r ) = − ∫ E • dl = −∫ Er ( dr ) = −∫ Er ( dr ) − ∫ Er ( dr ) = 4πε0 r =∞ ∞ ∞ a r =r
R
q Q +0 aR
ELECTRIC POTENTIAL for Charged Sphere (Y&F, ex.23.8) Suppose we have a charged metal sphere with charge q. What is the electric potential as a function of radius r?
N.B. V is continuous but E is not.
q
Clicker exercise 1
A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. – What is the value of the potential Va at
a Q
b
the inner surface of the spherical shell?
(a) Va = 0
(b)
Va =
1 4 πε
0
Q a
(c)
Va =
1 4 πε
0
Q b
Eout
Clicker exercise 1
A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. – What is the value of the potential Va at
a Q
b
the inner surface of the spherical shell?
(a) V a = 0 (b)
Va =
1 4 πε
0
Q a
(c)
Va =
1 4 πε
0
Q b
• How to start?? The only thing we know about the potential is a r r its definition:
Va ≡ Va − V∞ = −∫ E • dl ∞
• To calculate
Va, we need to know the electric field E
• Outside the spherical shell: • Apply Gauss’ Law to sphere: • Inside the spherical shell:
r a r r b r Va = − ∫ E • dl − ∫ E • dl ∞ b
r Q rˆ E= 4πε 0 r 2
E=0 1 Q Va = +0 4πε 0 b
Va =
1 4 πε
0
Q b
Preflight 6:
Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B.
A
B
2) Compare the potential at the surface of conductor A with the potential at the surface of conductor B. a) VA > VB
b) VA = VB
c) VA < VB
Electrical Potential Two ways to find V at any point in space: r r r • Use electric field: V = − ∫ E • dl ∞
• Sum or Integrate over charges: q1 q2 r2
P
r1 r3 q3
dq
r
1
qi V =∑ i 4πε 0 ri
P
V=
dq 4πε 0 ∫ r 1
Examples of integrating over a distribution of charge: • line of charge (review this one) • ring of charge • disk of charge You should be able to do these.
Infinite line charge or conducting cylinder. r
Linear charge density λ
λ E= 2πε 0 r b
Va − Vb = ∫ Edr = ∫ a Suppose we set rb to infinity, potential is infinite
rb λ λ = ln( ) 2πε 0 r 2πε 0 ra Instead, set ra=r and rb=r0 at some fixed radius r0.
Cover homework hints
Potential from a charged sphere 1 q V (r) = 4πε0 r
(where V (∞) ≡ 0 )
Er
Equipotential • • •
•
The electric field of the charged sphere has spherical symmetry. The potential depends only on the distance from the center of the sphere, as is expected from spherical symmetry. Therefore, the potential is constant on a sphere which is concentric with the charged sphere. These surfaces are called equipotentials. Notice that the electric field is perpendicular to the equipotential surface at all points.
Equipotentials Defined as: The locus of points with the same potential. •
Example: for a point charge, the equipotentials are spheres centered on the charge.
The electric field is always perpendicular to an equipotential surface!
r r VB − VA = − ∫ E • dl B
Why??
A
Along the surface, there is NO change in V (it’s an equipotential!) Therefore,
r r − ∫ E • dl = ∆V = 0 B
A
r r We can conclude then, that E • dl is zero. If the dot product of the field vector and the displacement vector is zero, then these two vectors are perpendicular, or the electric field is always perpendicular to the equipotential surface.
EXAMPLES of Equipotential Lines
Conductors +
+
+
+
+
+
+ +
+
+
+
+
+ +
• Claim The surface of a conductor is always an equipotential surface (in fact, the entire conductor is an equipotential). • Why?? If surface were not equipotential, there would be an electric field component parallel to the surface and the charges would move!!
Preflight 6:
B
A
3) The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now ? a) VA > VB
b) VA = VB
c) VA < VB
4) What happens to the charge on conductor A after it is connected to conductor B ? a) QA increases b) QA decreases c) QA doesn’t change
Charge on Conductors? • How is charge distributed on the surface of a conductor? – KEY: Must produce E=0 inside the conductor and E normal to the surface .
Spherical example (with little off-center charge): + + + - -- + - + + -+q - + + - + + - + + + + + +
E=0 inside conducting shell. charge density induced on inner surface non-uniform. charge density induced on outer surface uniform
E outside has spherical symmetry centered on spherical conducting shell.
Equipotential Example • Field lines more closely spaced near end with most curvature – higher E-field • Field lines ⊥ to surface near the surface (since surface is equipotential). • Near the surface, equipotentials have similar shape as surface. • Equipotentials will look more circular (spherical) at large r.
Conservation of Energy • The Coulomb force is a CONSERVATIVE force (i.e., the work done by it on a particle which moves around a closed path returning to its initial position is ZERO.) • Therefore, a particle moving under the influence of the Coulomb force is said to have an electric potential energy defined by: U = qV
this “q” is the “test charge” in other examples...
• The total energy (kinetic + electric potential) is then conserved for a charged particle moving under the influence of the Coulomb force.
Lecture 6, ACT 3 3A
Two test charges are brought separately to the vicinity of a positive charge Q. Q
r q
– charge +q is brought to pt A, a distance r from Q. – charge +2q is brought to pt B, a Q distance 2r from Q. – Compare the potential energy of q (UA) to that of 2q (UB):
(b) UA = UB
(a) UA < UB 3B
A B
2r
2q
(c) UA > UB
• Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r = ∞?
(a)
vf =
1 4 πε
0
Qq mr
(b)
v
f
=
1 2 πε
0
Qq mr
(c) v
f
= 0
Clicker Problem 3A
• Two test charges are brought separately to the vicinity of positive charge Q. – charge +q is brought to pt A, a distance r from Q. – charge +2q is brought to pt B, a distance 2r from Q.
Q
r
q A
Q
2r
– Compare the potential energy of q (UA) to that of 2q (UB):
(a) UA < UB
(b) UA = UB
(c) UA > UB
• The potential energy of q is proportional to Qq/r. • The potential energy of 2q is proportional to Q(2q)/(2r). • Therefore, the potential energies UA and UB are EQUAL!!!
2q B
Clicker Problem 3B
• Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r = ∞?
(a)
vf =
1
Qq 4πε 0 mr
(b)
vf =
1
Qq 2πε 0 mr
(c) v
= 0
f
• What we have here is a little combination of 170 and 272. • The principle at work here is CONSERVATION OF ENERGY. • Initially: • The charge has no kinetic energy since it is at rest. • The charge does have potential energy (electric) = UB. • Finally: • The charge has no potential energy (U ∝ 1/R) • The charge does have kinetic energy = KE
UB = KE
1 4 πε
0
Q( 2q ) 1 = mv 2r 2
2 f
v 2f =
1 2 πε
0
Qq mr
How to obtain vector E field from V r r V = − ∫ E • dl r
• Determine V from E:
∞
Example: V due to spherical charge distribution. Determining E from V:
r r b ∆V = Vb − Va = − ∫ E • dl = ∫ dV b
a
For an infinitesimal step:
r r dV = − E • dl = − E dl cos θ
a
E θ dl
Cases: • θ = 0: • θ = 90o: • θ = 180o:
directional derivative dV = E dl (maximum) dV = 0 dV depends on direction dV = -E dl
Can write:
r r dV = − E • dl = −( E x iˆ + E y ˆj + E z kˆ) • (dxiˆ + dy ˆj + dz kˆ) = −( E x dx + E y dy + E z dz )
Potential Gradient Take step in x direction: (dy = dz = 0)
dV = − ( E x dx + E y dy + E z dz ) = − E x dx dV ∂V Ex = − =− partial derivative dx y , z const . ∂x Similarly:
Ey = − And:
∂V ∂y
Ez = −
∂V ∂z
r r ∂V ˆ ∂V ˆ ∂V ˆ E = −( i+ j+ k ) = −∇V ∂x ∂y ∂z r ∂ ˆ ∂ ˆ ∂ ˆ gradient operator ∇ =( i + j + k) ∂x ∂y ∂z
Gradient of V points in the direction that V increases the fastest with respect to a change in x, y, and z. E points in the direction that V decreases the fastest. E perpendicular to equilpotential lines.
Potential Gradient Example: charge in uniform E field E
U = qEy
q
V = U/q = Ey where V is taken as 0 at y = 0.
r r ∂ ˆ ∂ ˆ ∂ ˆ E = −∇V = − ( i + j + k ) Ey ∂x ∂y ∂z = −(0 iˆ + E ˆj + 0kˆ) = − E ˆj Given E or V in some region of space, can find the other.
y
o
Cylindrical and spherical symmetry cases: Example: E of point charge: ∂V ∂ q For E radial case and r is distance Er = − =− ( ) from point (spherical) or axis ∂r ∂r 4πε 0 r (cylindridal):
∂V Er = − ∂r
= −(
q
4πε 0
)(
q −1 ) = r2 4πε 0 r 2
r r ∂ ∂ ˆ ∂ ˆ E = −∇V = − ( iˆ + j + k ) { A( x 2 − 3 y 2 + z 2 ) )} ∂z ∂x ∂y = −(2 Ax iˆ − 6 Ay ˆj + 2 Azkˆ) = −2 A( x iˆ − 3 yˆj + zkˆ)
UI7PF7: Clicker Problem
This graph shows the electric potential at various points along the x-axis.
2) At which point(s) is the electric field zero? A
B
C
D
UI7ACT1: Clicker Problem 1
The electric potential in a region of space is given by
V (x ) = 3x 2 − x 3 The x-component of the electric field Ex at x = 2 is (a)
Ex = 0
(b)
Ex > 0
(c)
Ex < 0
UI7ACT1: Clicker Problem 1
The electric potential in a region of space is given by
V ( x ) = 3x 2 − x 3 The x-component of the electric field Ex at x = 2 is (a)
Ex = 0
(b)
(c)
Ex > 0
We know V(x) “everywhere” To obtain Ex “everywhere”, use
Ex < 0
r r E = −∇ V ∂V Ex = − ∂x
E x = −6 x + 3x 2
E x (2 ) = −12 + 12 = 0
The Bottom Line/Take Home Message If we know the electric field E everywhere, W V B − V A ≡ AB q0
⇒
r r VB − V A = − ∫ E • dl B
A
allows us to calculate the potential function V everywhere (keep in mind, we often define VA = 0 at some convenient place)
If we know the potential function V everywhere, r r E = −∇ V allows us to calculate the electric field E everywhere
• Units for Potential! 1 Joule/Coul = 1 VOLT (but remember we measure potential differences with a voltmeter)