HIGHER EDU PRESENTATION

chapter IIIeLECTRIC POTENTIAL

LEARNING OUTCOME

• By the end of this lecture, you will be able to:
• Define the work done by an electric force.
• Define electric potential energy.
• Apply work and potential energy in systems with electric charges.
• Define electric potential, voltage, and potential difference.
• Define the electron-volt.
• Calculate electric potential and potential difference from potential energy and electric field.
• Apply conservation of energy to electric systems.

ELECTRICAL POTENTIAL ENERGY

ELECTRICAL POTENTIAL ENERGY, CONTINUED

ELECTRIC POTENTIAL

(5)

ELECTRIC POTENTIAL CONT.

Potential difference between points A and B in an electric field is

(5)

(6)

displacement between two points in space

WORK AND ELECTRIC POTENTIAL

• Assume a charge moves by external agent without any change kinetic energy.

• The work performed on the charge at constant velocity

SI Units for electric potential 1 V ≡ 1 J/C

• In addition, the electric field can be expressed 1 N/C = 1 V/m

• electric field is a measure of the rate of change of the electric potential with respect to position.

(6)

(7)

(8)

ELECTRON -VOLTS

• One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt.

• 1 eV = 1.60 x 10-19 J

Potential Difference in a Uniform Field

Fig 1

Potential Difference in a Uniform Field

(6)

(7)

Potential and the Direction of Electric Field

• When a positive test charge moves from A to B

• If q is positive, then ΔU is negative
• the electric potential energy decreases (loss) when q moves in direction of electric field

(5)

(7)

(8)

More About Directions

ΔV=ΔU q ΔU= qΔV= -qEd

• A system consisting of a negative charge and an electric field gains potential energy when the charge moves in the direction of the field.

• If negative charge release from rest in electric field, it accelates in opposite direction of the field

• To move in the direction of the electric field, an external agent must do positive work on the charge.

EQUIPOTENTIAL

• Eq (6) gives

(8)

(9)

All points in a plane perpendicular to uniform electric field --> exhibits the same electric potential VB = VC

Fig. 2

Equipotential surface is surface consist of a continuous distribution of points having the same electric potential

CONCEPT TEST

Which requires the most work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. A. P ->1 B. P->2 C. P->3 D. P->4 E. All require the same amount of work

CONCEPT TEST

Which requires the most work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. A. P ->1 B. P->2 C. P->3 D. P->4 E. All require the same amount of work

example: Electric field between two parallel plate

A battery has a specified potential difference between its terminal and establishes that potential difference between conductors attached to the terminals. A 12V battery is connected between two parallel plates as shown in Fig. 3. The separation between plates is d= 0.30 m, and we assume the electric field between the plates is uniform. Find the magnitude of the electric field between the plates.

Fig. 3

POTENTIAL AND POINT CHARGES

Charged Particle in a Uniform Field, Example

• It is customary to choose a reference potential of V = 0 at rA = ∞.

• Then the potential due to a point charge at some point r is:

• The electric potential due to several point charges is the sum of the potentials due to each individual charge.

• The sum is the algebraic sum

POTENTIAL ENERGY OF MULTIPLE CHARGES

U with Multiple Charges, final

Concept test

A +Q charged particle is fixed at the origin. You are asked to bring another particle with charge +Q from a point A, 5 m from the fixed particle, to a point B, 2 m from the fixed particle. Do you do positive, negative, or zero work on the particle to move it from point A to point B? A. Positive B. Negative C. Zero D. The sign of the work cannot be determined unless the path between A and B is given.

Concept test

A +Q charged particle is fixed at the origin. You are asked to bring another particle with charge +Q from a point A, 5 m from the fixed particle, to a point B, 2 m from the fixed particle. Do you do positive, negative, or zero work on the particle to move it from point A to point B? A. Positive B. Negative C. Zero D. The sign of the work cannot be determined unless the path between A and B is given.

CONCEPT TEST

A +Q charged particle is fixed at the origin. You are asked to bring another particle with charge +Q from a point A, 5 m from the fixed particle, to a point B, 2 m from the fixed particle. Does the system of two charges gain or lose electric potential energy in this process? A. Gain energy B. Lose energy C. No change in energy D. The sign of the energy change cannot be determined unless the path between A and B is given.

CONCEPT TEST

A +Q charged particle is fixed at the origin. You are asked to bring another particle with charge +Q from a point A, 5 m from the fixed particle, to a point B, 2 m from the fixed particle. Does the system of two charges gain or lose electric potential energy in this process? A. Gain energy B. Lose energy C. No change in energy D. The sign of the energy change cannot be determined unless the path between A and B is given.

CONCEPT TEST

A +Q charged particle is fixed at the origin. You are asked to bring another particle with charge -Q from a point A, 5 m from the fixed particle, to a point B, 2 m from the fixed particle. Does the system of two charges gain or lose electric potential energy in this process? A. Gain energy B. Lose energy C. No change in energy D. The sign of the energy change cannot be determined unless the path between A and B is given.

CONCEPT TEST

A +Q charged particle is fixed at the origin. You are asked to bring another particle with charge -Q from a point A, 5 m from the fixed particle, to a point B, 2 m from the fixed particle. Does the system of two charges gain or lose electric potential energy in this process? A. Gain energy B. Lose energy C. No change in energy D. The sign of the energy change cannot be determined unless the path between A and B is given.

EXAMPLE

A charge q1 =2 uC is located at the origin and a charge q2 = -6 uC is located at (0, 3.00) m. (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. Solution

CONCEPT TEST

What is the electric potential at point A?

A. V > 0 B. V = 0 C. V < 0

CONCEPT TEST

What is the electric potential at point A?

A. V > 0 B. V = 0 C. V < 0

CONCEPT TEST

What is the electric potential at point B?

A. V > 0 B. V = 0 C. V < 0

CONCEPT TEST

What is the electric potential at point B?

A. V > 0 B. V = 0 C. V < 0

Finding E From V

Assume, to start, that the field has only an x component. Similar statements would apply to the y and z components. Equipotential surfaces must always be perpendicular to the electric field lines passing through them.

E and V for an Infinite Sheet of Charge, dipole and point charge

The electric field is radial. Er = - dV / dr

The equipotential lines are the dashed blue lines.The electric field lines are the brown lines. The equipotential lines are everywhere perpendicular to the field lines.

Electric Field from Potential, General

In general, the electric potential is a function of all three dimensions. Given V (x, y, z) you can find Ex, Ey and Ez as partial derivatives:

Electric Potential for a Continuous Charge Distribution

Method 1:

• The charge distribution is known.
• Consider a small charge element dq
• Treat it as a point charge.
• The potential at some point due to this charge element is

Electric Potential for a Continuous Charge Distribution

• To find the total potential, you need to integrate to include the contributions from all the elements.

• This value for V uses the reference of V = 0 when P is infinitely far away from the charge distributions.

V for a Continuous Charge Distribution, final

• If the electric field is already known from other considerations, the potential can be calculated using the original approach:

• If the charge distribution has sufficient symmetry, first find the E from Gauss’ Law and then find the potential difference between any two points,

• Choose V = 0 at some convenient point

Problem-Solving Strategies for Calculating Electric Potential

Conceptualize

• Think about the individual charges or the charge distribution.
• Imagine the type of potential that would be created.
• Appeal to any symmetry in the arrangement of the charges.

Categorize

• Group of individual charges or a continuous distribution?
• The answer will determine the procedure to follow in the analysis step.

Problem-Solving Strategies 2

Analyze General

• Scalar quantity, so no components
• Use algebraic sum in the superposition principle
• Keep track of signs
• Only changes in electric potential are significant
• Define V = 0 at a point infinitely far away from the charges.
• If the charge distribution extends to infinity, then choose some other arbitrary point as a reference point.

Problem-Solving Strategies 3

Analyze, cont

• If a group of individual charges is given
• Use the superposition principle and the algebraic sum.

• If a continuous charge distribution is given
• Use integrals for evaluating the total potential at some point.
• Each element of the charge distribution is treated as a point charge.

• If the electric field is given
• Start with the definition of the electric potential.
• Find the field from Gauss’ Law (or some other process) if needed.

Problem-Solving Strategies, Final

Finalize

• Check to see if the expression for the electric potential is consistent with your mental representation.
• Does the final expression reflect any symmetry?
• Image varying parameters to see if the mathematical results change in a reasonable way.

V for a Uniformly Charged Ring

P is located on the perpendicular central axis of the uniformly charged ring. The symmetry of the situation means that all the charges on the ring are the same distance from point P.

• The ring has a radius a and a total charge Q.

The potential and the field are given by

V for a Uniformly Charged Disk

The ring has a radius R and surface charge density of σ. P is along the perpendicular central axis of the disk. P is on the central axis of the disk, symmetry indicates that all points in a given ring are the same distance from P. The potential and the field are given by

* refer to video in elearn for detail explanation

V for a Finite Line of Charge

A rod of line ℓ has a total charge of Q and a linear charge density of λ.

• There is no symmetry to use, but the geometry is simple.

* refer to video in elearn for detail explanation

V Due to a Charged Conductor

V Due to a Charged Conductor, cont.

V is constant everywhere on the surface of a charged conductor in equilibrium.

• ΔV = 0 between any two points on the surface

The surface of any charged conductor in electrostatic equilibrium is an equipotential surface. Every point on the surface of a charge conductor in equilibrium is at the same electric potential. Because the electric field is zero inside the conductor, we conclude that the electric potential is constant everywhere inside the conductor and equal to the value at the surface.

Cavity in a Conductor

Assume an irregularly shaped cavity is inside a conductor. Assume no charges are inside the cavity. The electric field inside the conductor must be zero.

Cavity in a Conductor

The electric field inside does not depend on the charge distribution on the outside surface of the conductor. For all paths between A and B, A cavity surrounded by conducting walls is a field-free region as long as no charges are inside the cavity.

Corona Discharge

• If the electric field near a conductor is strong, electrons due to random ionizations of air molecules accelerate away from their parent molecules.

• These electrons ionize additional molecules near the conductor creating more free electrons.

• The corona discharge is the glow that results from the recombination of these free electrons with the ionized air molecules.

• If a conductor has an irregular shape, the electric field can be very high near sharp points or edges of the conductor

Corona Discharge

• Used to locate broken or faulty components.

• A broken insulator on a transmission tower has sharp edges where corona discharge likely to occur.

• Occur at the sharp end of a broken conductor strand.

Electrical corona effect

Millikan oil-drop experiment

• Robert Millikan measured e, the magnitude of the elementary charge of the electron.
• Oil droplets pass through a small hole and are illuminated by a light.

MILLIKAN oil-drop experiment

With no electric field between the plates, the gravitational force and the drag force (viscous) act on the electron. The drop reaches terminal velocity with

MILLIKAN oil-drop experiment

An electric field is set up between the plates. The upper plate has a higher potential. The drop reaches a new terminal velocity when the electrical force equals the sum of the drag force and gravity.

MILLIKAN oil-drop experiment

The drop can be raised and allowed to fall numerous times by turning the electric field on and off. After many experiments, Millikan determined:

• q = ne where n = 0, -1, -2, -3, …
• e = 1.60 x 10-19 C

This yields conclusive evidence that charge is quantized.

MILLIKAN OIL-DROP EXPERIMENT

ELECTROSTATIC PRECIPITATOR

An application of electrical discharge in gases is the electrostatic precipitator. It removes particulate matter from combustible gases. The air to be cleaned enters the duct and moves near the wire. As the electrons and negative ions created by the discharge are accelerated toward the outer wall by the electric field, the dirt particles become charged. Most of the dirt particles are negatively charged and are drawn to the walls by the electric field.

Van de Graaff Generator

Charge is delivered continuously to a high-potential electrode by means of a moving belt of insulating material. The high-voltage electrode is a hollow metal dome mounted on an insulated column. Large potentials can be developed by repeated trips of the belt. Protons accelerated through such large potentials receive enough energy to initiate nuclear reactions.

THE VAN DE GRAAFF GENERATOR

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