Physics 2054c spring 2002
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Physics 2054C – Spring 2002. Review for Quiz on Wed. Start Chapter 17. Housekeeping. Lab Sections. Most People Assigned Let me know if you are in a lab section and need to change. Provide a 1 st and 2 nd choice. CAPA. Need names of those without CAPA logins. Other Questions?.

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Physics 2054C – Spring 2002

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Physics 2054c spring 2002

Physics 2054C – Spring 2002

Review for Quiz on Wed.

Start Chapter 17

Larry Dennis


Housekeeping

Housekeeping

  • Lab Sections.

    • Most People Assigned

    • Let me know if you are in a lab section and need to change.

    • Provide a 1st and 2nd choice.

  • CAPA.

    • Need names of those without CAPA logins.

  • Other Questions?


Electric forces coulomb s law

F = kQ1Q2

R2

Electric Forces (Coulomb’s Law)

  • Like Charges Repel.

  • Opposite Charges Attract.

-

+


Coulomb s law newton s law

-

F = kQ1Q2 = ma

R2

+

Coulomb’s Law & Newton’s Law

  • Both apply

  • Electrical forces cause accelerations.

  • It is a vector.


Electric fields

Electric Fields

  • Related to the Electric Force.

  • Removes the dependence on one of the charges.

  • E = F/Q1 = kQ2/R2


Rules for drawing electric field lines

Rules for Drawing Electric Field Lines

  • The lines must begin on a positive charges and terminate on a negative charges (or at infinity).

  • The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.

  • No two field lines can cross.


Superposition

Superposition

  • Add the effects of multiple charges.

  • Fnet = F1 + F2 + F3 .

    • Calculate each of the forces as if the other charges were not present.

  • Enet = E1 + E2 + E3 .

    • Calculate each of the electric fields as if the other charges were not present.


Sample problem

L

R

0.6 m

Sample Problem

  • Two charges are sitting on a table separated by a distance of 0.60 m. The sign and magnitude of these two charges (L and R) are unknown. We are going to use a positive test charge (Q = 5.0 x 10-6 C) to determine which charge is positive and which charge is negative.

    • If the charges L and R repel then they must both positive or both be negative.

    • If the charges attract (the case shown below), then they must be of opposite sign.

    • For now we will assume the attract one another.


Sample problem continued

Q

L

R

0.6 m

Sample Problem (continued)

  • We place our test charge Q midway between the two charges and observe that it moves to the right (towards R). From this we can determine that:

  • Q could be opposite R (attract), same as L (repel). Or

  • R could be much less than L and both are positive, but these charges would repel one another. Or

  • L could be much less than R and both are negative, but again, L and R would repel one another.

Case 1 is our only choice.


Sample problem continued1

This is the one!

L

R

Case 1: R negative & L positive

L

R

Case 2: R positive & L negative

Sample Problem (continued)

  • What is the direction of the electric field at the midpoint?

    • (recall) F = Q E and since Q>0, F and E are parallel.

    • Because the positive charge, Q, moved to the right, the electric field must be to the right.


Sample problem continued2

Q

R

0.3 m

Sample Problem (continued)

  • Suppose that we remove charge L and measure the force exerted on Q by R to be 8.0 N to the right.

    • What is the sign of R?

    • What is the sign of L?

R must be negative since it attracts a positive charge!

L must be positive since it must be opposite to R!


Sample problem continued3

Q

R

0.3 m

Sample Problem (continued)

  • What is the magnitude of the electric field at the point where Q is located due to charge R?

F = Q*ER

ER = FR/Q

= 8.0 N / 5.0 x 10-6 C

= 1.6 x 106 N/C


Sample problem continued4

Q

R

0.3 m

Sample Problem (continued)

  • What is the magnitude of the charge R?

ER = kQR/D2

QR = ER * D2 /k

= 1.6 x 106 N/C * (0.3 m)2 /9.0 x 109 N-m2 /C2

QR = 16 x 10-6 C (16 microcoulombs)


Chapter 17 electrical energy and electrostatic potential

Chapter 17 – Electrical Energy and Electrostatic Potential

  • Electric forces can be used to change the energy of matter.

    • Wab = -qVab = -q(Va – Vb)

    • Electric potential (often shortened to potential) is the change in potential energy per unit charge due to an electric field.

    • Measured in volts.

    • -Wab/q = (Va – Vb)


Electric potential energy

Electric Potential Energy

  • PE = PEb = PEa = q(Vb – Va)

Vab

-

+

W = qV

W = Fd

W = (qE)d

W = qV = qEd

 E = V/d

+q

d


Electric potential energy1

Vab = 1500 V

-

+

+q = .5 C

d = 2 mm

Electric Potential Energy

  • PE = PEb = PEa = q(Vb – Va)

W = qV = .5 C * 1500 = 750 J

W = qV = qEd

 E = V/d = 1500 V / 0.002 m

= 5.5 x 105 V/m = N/C


The electron volt

The Electron Volt

  • Energy acquired by an electron as an electron moves through 1 volt.

Vab

-

+

W = qV

W = 1.602 x 10-19 C * 1 Volt

W = 1.602 x 10-19 Joules

W = 1 eV (electron volt)

+q

d


Next time

Next Time

  • Tutorials:

    • Mon. 12:15 – 3:15

    • Tue. 9:00 – 12:00

  • CAPA Due Wednesday Morning at 2:00 AM

  • Wed: First Quiz on Chapter 17

  • Chapter 17 - Electrical Energy


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