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Electric and 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 to pass through the rod then use a small radius of curvature.

Air is normally an , however for large E fields (E>3 x 10 6 V/m) it starts conducting.

Empire State Building, NYC Electrical Potential Review: Wa → b = done by in going from a to b along path. b b r r b r r Wa →b = F • ld = qE • ld ∫a ∫a F b r r θ ∆U = Ub −U a = −Wa →b = −∫ qE • ld a a dl U =

b r r ∆U Ub −U a Wa →b ∆V = Vb −Va = = = − = −∫ E • ld q q q a

• Potential difference is minus the work done per unit charge by the electric 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 r V = V = −∫ E • ld ∞ allows us to calculate V everywhere if we know E Potential from charged spherical shell

• E-field (from Gauss' Law) V q q • : 4πεπεπε 000 R 4πεπεπε 000 r r < R Er = 0 1 q • r >R: E = R r r 4πε 2 0r R

• Potential R • r > R:

r === r r r r 1 Qq V r( ) === −−− ∫∫∫ E ••• ld === −−− ∫∫∫ E r ( dr ) === r === ∞∞∞ ∞∞∞ 4πε 0 πεr • r < R: r===r r r r aR r 1 Qq V )r( === −−− ∫∫∫ E••• ld === −−−∫∫∫ Er dr( )=== −−−∫∫∫ Er dr( )−−−∫∫∫ Er dr( ) === +++0 4πε0 πεaR r===∞∞∞ ∞∞∞ ∞∞∞ Ra ELECTRIC POTENTIAL for Charged Sphere (Y&F, ex.23.8)

Suppose we have a charged q 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. Clicker exercise

1 A point charge Q is fixed at the a center of an uncharged conducting Q spherical shell of inner radius a and outer radius b. b – What is the value of the potential Va at the inner surface of the spherical shell? 1 Q 1 Q V = V = (a) Va = 0 (b) a (c) a 4 πε 0 a 4πε 0 b E Clicker exercise out 1 A point charge Q is fixed at the center of an uncharged conducting a spherical shell of inner radius a Q and outer radius b. – What is the value of the potential at Va b the inner surface of the spherical shell? 1 Q 1 Q V a = V = (a) Va = 0 (b) 4πε a (c) a 0 4πε 0 b • How to start?? The only thing we know about the potential is its definition: a r r Va ≡Va −V∞ = −∫ E • ld ∞

• To calculate Va, we need to know the E • Outside the spherical shell: r Q rˆ E = • Apply Gauss’ Law to sphere: 2 4πε0 r • Inside the spherical shell: E = 0 b r r a r r 1 Q 1 Q V = V = − ∫ E • ld − ∫ E • ld Va = + 0 a a 4πε 0 b ∞ b 4πε0 b Preflight 6:

Two spherical conductors are separated by a large . 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 > V B b) VA = V B c) VA < V B Electrical Potential Two ways to find V at any point in : r r r • Use electric field: V = − E • ld ∫∞ • Sum or Integrate over charges:

q 1 r P 1 q 1 = i q2 r2 V ∑ i 4πε0 ri r3 q 3 P r 1 dq dq V = ∫ 4πε0 r

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 λ

λ E = 2πε0 r b λλ rb Va−= V b ∫ Edr = ∫ = ln( ) a 2πε πε0r 2 0 r a

Suppose we set rb to infinity, Instead, set ra=r and rb=r 0 at potential is infinite some fixed radius r 0. Cover homework hints Potential from a charged sphere

1 q V( r ) = 4πε0 r E (where V ( ∞ ) ≡ 0 ) r

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 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!

B r r Why?? VB −VA = −∫ E • ld A Along the surface, there is NO change in V (it’s an equipotential!) B r Therefore, r − ∫ E • ld = ∆V = 0 A r r We can conclude then, that E • ld is zero.

If the of the field vector and the 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:

A B

3) The two conductors are now connected by a wire. How do the at the conductor surfaces compare now ?

a) VA > V B b) VA = V B c) VA < V B

4) What happens to the charge on conductor A after it is connected to conductor B ?

a) QA increases

b) Q A decreases

c) Q A 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):

E=0 inside conducting shell. + + + + - - + - - - induced on + - + inner surface non-uniform. - +q - + - - + - - + 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.

• The force is a (i.e., the work done by it on a which moves around a closed path returning to its initial 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 Two test charges are brought 3A separately to the vicinity of a positive charge Q. Q r – charge +q is brought to pt A, a q distance r from Q. A – charge +2 q is brought to pt B, a distance 2r from Q. Q 2r B – Compare the potential energy of q 2q (UA) to that of 2q (UB):

(a) UA < UB (b) UA = UB (c) UA > UB

3B • Suppose charge 2q has m and is released from rest from the above position (a distance 2r from Q).

What is its vf as it approaches r = ∞?

1 Qq v = 1 Qq (c) (a) f (b) v f = v f = 0 4πε 0 mr 2πε 0 mr Clicker Problem • Two test charges are brought 3A separately to the vicinity of positive charge Q. r Q q – charge +q is brought to pt A, a A distance r from Q. Q 2q – charge +2 q is brought to pt B, a 2r distance 2r from Q. B

– 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)/(2 r).

• Therefore, the potential UA and UB are EQUAL!!! 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 = ∞?

1 Qq 1 Qq (a) v f = (b) v f = (c) v f = 0 4πε0 mr 2πε 0 mr

• 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 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

1 Q( 2q ) 1 2 1 Qq = mv 2 v = U =KE f f B 4πε 0 2r 2 2πε 0 mr How to obtain vector E field from V r r r • Determine V from E: V = − E • ld ∫∞ Example: V due to spherical charge distribution.

Determining E from V: b r r b ∆V = Vb −Va = − E • ld = dV ∫a ∫a For an step: E r r θ dV = −E • ld = −Edl cos θ dl Cases: directional • θ = 0: dV = E dl (maximum) • θ = 90 o: dV = 0 • θ = 180 o: dV = -E dl dV depends on direction Can write: r r ˆ ˆ ˆ ˆ ˆ ˆ dV = −E • ld = −(Exi + Ey j + Ez k) • (dx i + dy j + dz k)

= −(Ex dx + Ey dy + Ez dz ) Potential Take step in x direction: (dy = dz = 0)

dV = − (Ex dx + Ey dy + Ez dz ) = −Ex dx dV ∂V Ex = − = − dx y,z const . ∂x Similarly: ∂V ∂V Ey = − E = − ∂y z ∂z And: r ∂V ∂V ∂V r E = −( iˆ + ˆj + kˆ) = −∇V ∂x ∂y ∂z r ∂ ∂ ∂ ∇ =( iˆ + ˆj + kˆ) gradient operator ∂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.

Example: charge in uniform E field U = qEy E

V = U/q = Ey q where V is taken as 0 at y = 0. r r ∂ ∂ ∂ E = −∇V = −( iˆ + ˆj + kˆ) Ey ∂x ∂y ∂z y = − 0( iˆ + E ˆj + 0kˆ) = −E ˆj Given E or V in some region of space, o can find the other.

Cylindrical and spherical symmetry cases: Example: E of point charge: ∂V ∂ q For E radial case and r is distance E = − = − ( ) from point (spherical) or axis r ∂r ∂r 4πε0 r (cylindridal): ∂V q −1 q = −( )( ) = Er = − 2 2 ∂r 4πε0 r 4πε0 r r r ∂ ∂ ∂ E = −∇V = −( iˆ + ˆj + kˆ {) A(x2 − 3y 2 + z )2 )} ∂x ∂y ∂z = − 2( Ax iˆ − 6Ay ˆj + 2Az kˆ) = −2A( ix ˆ − 3 ˆjy + kz ˆ) 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)= 3x2 − x3

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)= 3x2 − x3

The x-component of the electric field Ex at x = 2 is

(a) Ex = 0 (b) Ex > 0 (c) Ex < 0

We know V(x) “everywhere” r r E = −∇V To obtain Ex “everywhere”, use

∂V E = − x ∂x

2 Ex = −6x + 3x

Ex (2)= −12 +12 = 0 The Bottom Line/Take Home Message

If we know the electric field E everywhere, W B r r − ≡ AB V − V = − E • ld V B V A ⇒⇒⇒ B A ∫ q 0 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 /Coul = 1 (but remember we measure potential differences with a )