Ch – 30 Potential and Field

Ch – 30 Potential and Field

Learning Objectives – Ch 30

• To establish the relationship between and V.• To learn more about the properties of a

conductor in electrostatic equilibrium.• To introduce batteries as a practical source of

potential difference.• To find the connection between current and

potential difference for a conductor.• To find the connection between charge and

potential difference for a capacitor.• To analyze simple capacitor circuits.

Finding potential (V) from field (E)

-ΔU =

ΔV = ΔU/q, E =F/q , therefore

If you want to find a value for Vf , instead of ∆V, you must specify a position, si , where Vi =0. This position is often at infinity.

Finding E field from potential (V)

Es = - dV/ds

Given this graph of V, make a graph of Ex Assume the equation of the parabola is of

the form y = Kx2 where K is a constant

Answer

E = -dV/dx

for parabola K = 50,000 V/m2

E0-2cm =(-)100,000x

E2-4cm = 0

E4-6cm = (-)-20V/.02m

=1000V/m

Given this graph of Ex , make a graph of V (V = 0 at x = 0)

Answer

∆V = area under the curve (V0 = 0 at x=0)

∆ V1-3 = (-)-200V x .02m

= +4 V

∆ V3-4 = ∆ V1-3 = +4V

∆ V4-6 = (-) 200V x .02m

= -4 V

A point charge of + 5/9nC is located at the origin

• Determine the values of x at which the potential is 100, 200, 300, 400, 500V.

• Graph V vs x along an x axis with the charge at the origin.

• Describe E, the electric field on both sides of the point charge (e.g.positive, negative, constant, increasing, decreasing).

Graph of V vs x

V x(cm)

100 ± 5.0

200 ± 2.5

300 ± 1.67

400 ± 1.25

500 ± 1.0

0

100

200

300

400

500

600

-6 -4 -2 0 2 4 6

x (cm)

V

For values of x > 0, E is positive and decreasing with increasing value of x

For values of x < 0, E is negative and decreasing with increasing values of |x|

Geometry of Potential and Field

• The direction of the electric field is perpendicular to the equipotential surfaces

• E always points in the direction of decreasing potential

• Field strength (magnitude) of E is inversely proportional to the spacing ∆s between equipotential surfaces

Which set of equipotential surfaces is valid for the electric field shown?

Answer

• Answer is c• Field strength

(magnitude) of E is inversely proportional to the spacing ∆s between equipotential surfaces

Kirchoff’s Loop Law

• The sum of all potential differences encountered while moving around closed path is zero

• This is a result of the conservation of energy for a conservative force

Conductor in electrostatic equilibrium

• For a sphere of charge Q, outside the conductor (r >R), E = kQ/r2

with the maximum field strength at the surface of the sphere E = kQ/R2

• Inside the sphere, E=0.

Conductor in electrostatic equilibrium

• To find ∆V between 2 points outside the sphere, integrate E along a line between the two points.

• At the surface of the sphere, there is a nonzero potential.

Conductor in electrostatic equilibrium

• Inside, since E = 0, ∆V =0, which means the potential inside is constant

• When a conductor is in electrostatic equilibrium, the entire conductor has the same potential, not necessarily the same charge

Conductor in electrostatic equilibrium

• At the surface of the conductor

• Same charge and same potential here .

• Same potential, but different charge and different electric field here

Connecting Potential and Current

• I = AJ = A σ E• In the battery shown: Ewire = |∆Vwire |

LWe can derive:I = ∆Vwire

RWhere R = L / A σ

R = ρL / AR is a property of a specific

wire, depending on the material, length and area.

Current Ranking Task

• Rank the currents I1–I5 at the five labeled points in this figure, from greatest to least. Explain.

Fat and Skinny

• Which current is greater? Both wires are made of the same material.

• Explain.

2 conductive rods

• Two conductive rods have been connected to a 6 V battery for “a long time.” What are the values of:

• ∆V12 _______• ∆ V34 _______• ∆ V23_______

Equipotential Map

• Compare the field strengths Ea and Eb. Are they equal, or is one larger than the other? Explain.

• Compare the field strengths Ec and Ed. Are they equal, or is one larger than the other? Explain.

• Draw the electric field vectors at points a through e.

Answers

• Electric field strength is the gradient of the potential

• Ea > Eb

• Ec >Ed

E field vectors and potential

• Is the potential at point 1 greater than, less than or equal to the potential at point 2? Explain.

• Determine a value of ∆V12 (1 is initial position, 2 is final position).

• Draw a series of equipotential surfaces spaced every 5 V.

Answer

• E points in direction of decreasing potential so V1 > V2

• |∆V12 |= 200 V/m x 10 cm = 20 V and since it is decreasing:

∆V12 = -20V• Arbitrary assumption

that V2 = 0 for equipotentials

0V

10V

20V

What is the Value for E?

• Assume V is a linear function as in a capacitor.

• Draw an arrow on the figure showing direction.

Answer

• E = 1000 V/m

Answer