Wednesday, Sep. 14, 2011 1 PHYS 1444-04 Dr. Andrew Brandt PHYS 1444 – Section 04 Lecture #5...

Wednesday, Sep. 14, 2011 1PHYS 1444-04 Dr. Andrew Brandt

PHYS 1444 – Section 04Lecture #5

• Chapter 21: E-field examples • Chapter 22: Gauss’ Law Examples• Chapter 23:

- Electric Potential Energy-Electric Potential

Wednesday Sep. 14, 2011Dr. Andrew Brandt

Homework on Ch 22 due Tuesday 20th at 9pm

Example 21-7Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x 10-31 kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?

Solution: a. The electric fields add in magnitude, as both are directed towards the negative charge. E = 6.3 x 108 N/C. b. We don’t know the relative lengths of E1 and E2 until we do the calculation. The acceleration is the force (charge times field) divided by the mass, and will be opposite to the direction of the field (due to the negative charge of the electron). Substitution gives a = 1.1 x 1020 m/s2.

21-7 Electric Field Calculations for Continuous Charge Distributions

A continuous distribution of charge may be treated as a succession of infinitesimal (point) charges. The total field is then the integral of the infinitesimal fields due to each bit of charge:

Remember that the electric field is a vector; you will need a separate integral for each component.

21-7 Electric Field for Continuous Charge DistributionsExample 21-9: A ring of charge.

A thin, ring-shaped object of radius a holds a total charge +Q distributed uniformly around it. Determine the electric field at a point P on its axis, a distance x from the center. Let λ be the charge per unit length (C/m).

Solution: Because P is on the axis, the transverse components of E must add to zero, by symmetry. The longitudinal component of dE is dE cos θ, where cos θ = x/(x2 + a2)1/2. Write dQ = λdl, and integrate dl from 0 to 2πa. Answer: E = (1/4πε0)(Qx/[x2 + a2]3/2)

Example 21-8Calculate the total electric field (a) at point A and (b) at point B in the figure due to both charges, Q1 and Q2.

Solution: The geometry is shown in the figure. For each point, the process is: calculate the magnitude of the electric field due to each charge; calculate the x and y components of each field; add the components; recombine to give the total field.a. E = 4.5 x 106 N/C, 76° above the x axis.b. E = 3.6 x 106 N/C, along the x axis.


Conceptual Example 21-10: Charge at the center of a ring.

Imagine a small positive charge placed at the center of a nonconducting ring carrying a uniformly distributed negative charge. Is the positive charge in equilibrium if it is displaced slightly from the center along the axis of the ring, and if so is it stable? What if the small charge is negative? Neglect gravity, as it is much smaller than the electrostatic forces.

Solution: The positive charge is in stable equilibrium, as it is attracted uniformly by every part of the ring. The negative charge is also in equilibrium, but it is unstable; once it is displaced from its equilibrium position, it will accelerate away from the ring.


Example 21-12: Uniformly charged disk.

Charge is distributed uniformly over a thin circular disk of radius R. The charge per unit area (C/m2) is σ. Calculate the electric field at a point P on the axis of the disk, a distance z above its center.

Solution: The disk is a set of concentric rings, and we know (from example 21-9) what the field due to a ring of charge is. Write dQ = σ 2πr dr. Integrate r from 0 to R. See text for answer.


Example 22 – 2 Flux from Gauss’ Law: Consider the two Gaussian surfaces, A1 and A2, shown in the figure. The only charge present is the charge +Q at the center of surface A1. What is the net flux through each surface A1 and A2?

• The surface A1 encloses the charge +Q, so from Gauss’ law we obtain the total net flux

• For A2 the charge, +Q, is outside the surface, so the total net flux is 0.

E dA rr

Ñ

E dA rr

Ñ

0

Q

0

00


Example 22 – 5 Long uniform line of charge: A very long straight wire possesses a uniform positive charge per unit length, . Calculate the electric field at points near but outside the wire, far from the ends.• Which direction do you think the field due to the charge on the wire is?

– Radially outward from the wire, the direction of radial vector r.• Due to cylindrical symmetry, the field is constant anywhere on the

Gaussian surface of a cylinder that surrounds the wire.– The end surfaces do not contribute to the flux at all. Why?

• Because the field vector E is perpendicular to the surface vector dA.

• From Gauss’ lawE dA

rrÑ

02E

r

Solving for E

E dAÑ 2E rl 0

enclQ

0

l

Example 22-4: Solid sphere of charge.

An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0).

Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2).b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r0

3 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r0

3).

Thursday, Mar. 3, 2011 10PHYS 1444-02 Dr. Andrew Brandt


Electric Potential Energy• Concept of energy is very useful solving mechanical problems• Conservation of energy makes solving complex problems easier. • When can the potential energy be defined?

– Only for a conservative force.– The work done by a conservative force is independent of the path. What

does it only depend on?? • The difference between the initial and final positions

– Can you give me an example of a conservative force?• Gravitational force

• Is the electrostatic force between two charges a conservative force?– Yes. Why?– The dependence of the force on distance is identical to that of the

gravitational force.• The only thing matters is the direct linear distance between the object not the

path.


Electric Potential Energy• What is the definition of change in electric potential energy Ub –Ua?

– The gain (or loss) of potential energy as the charge moves from point a to point b.– The negative work done on the charge by the electric force to move it from a to b.

• Let’s consider an electric field between two parallel plates w/ equal but opposite charges– The field between the plates is uniform since the gap is

small and the plates are infinitely long…

• What happens when we place a small charge, +q, on a point at the positive plate and let go?– The electric force will accelerate the charge toward

negative plate. – What kind of energy does the charged particle gain?

• Kinetic energy


Electric Potential Energy• What does this mean in terms of energies?

– The electric force is a conservative force.– Thus, the mechanical energy (K+U) is conserved

under this force.– The charged object has only electric potential

energy at the positive plate.– The electric potential energy decreases and – Turns into kinetic energy as the electric force works

on the charged object and the charged object gains speed.

• Point of greatest potential energy for– Positively charged object– Negatively charged object

PE=KE=ME=

U 00 KU

U+K=constK


Electric Potential• How is the electric field defined?

– Electric force per unit charge: F/q• We can define electric potential (potential) as

– The electric potential energy per unit charge– This is like the voltage of a battery…

• Electric potential is written with a symbol V– If a positive test charge q has potential energy Ua at

a point a, the electric potential of the charge at that point is a

a

UV

q


Electric Potential• Since only the difference in potential energy is meaningful,

only the potential difference between two points is measurable

• What happens when the electric force does “positive work”?– The charge gains kinetic energy– Electric potential energy of the charge decreases

• Thus the difference in potential energy is the same as the negative of the work, Wba, done on the charge by the electric field to move the charge from point a to b.

• The potential difference Vba is

baV b aV V b aU U

q

baW

q


• What does the electric potential depend on?– Other charges that create an electric field– What about the test charge?

• No, the electric potential is independent of the test charge• Test charge gains potential energy by existing in the potential created by other charges

• Which plate is at a higher potential?– Positive plate. Why?

• Since positive charge has the greatest potential energy.– What happens to the positive charge if it is let go?

• It moves from higher potential to lower potential– How about a negative charge?

• Its potential energy is higher on the negative plate. Thus, it moves from negative plate to positive. Potential difference is the same for

a negative charge at the negative plate as a positive charge at the positive plate.

• The unit of the electric potential is Volt (V).• From the definition, 1V = 1J/C.

Zero point of electric potential can be chosen arbitrarily.

Often the ground, a conductor connected to Earth is zero.

A Few Things about Electric Potential

Wednesday, Sep. 14, 2011 1 PHYS 1444-04 Dr. Andrew Brandt PHYS 1444 – Section 04 Lecture #5...

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