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

Review for Quiz on Wed.

Start Chapter 17

Larry Dennis

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?

F = kQ1Q2

R2

Electric Forces (Coulomb’s Law)
• Like Charges Repel.
• Opposite Charges Attract.

-

+

-

F = kQ1Q2 = ma

R2

+

Coulomb’s Law & Newton’s Law
• Both apply
• Electrical forces cause accelerations.
• It is a vector.
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
• 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
• 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.

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.

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.

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.

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!

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

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
• 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
• PE = PEb = PEa = q(Vb – Va)

Vab

-

+

W = qV

W = Fd

W = (qE)d

W = qV = qEd

 E = V/d

+q

d

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
• 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
• 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