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# Physics 2054C – Spring 2002 PowerPoint PPT Presentation

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?.

Physics 2054C – Spring 2002

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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