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### Descriptive Statements:

- Analyze phenomena related to static electricity, such as the behavior of electroscopes, induced molecular polarization, and charging by induction.
- Describe electric forces and electric fields for various simple charge distributions.
- Describe the motion of a charged particle in a constant electric field.
- Demonstrate knowledge of electric potential energy and potential difference.

### Sample Item:

Three identical point charges are located on the vertices of an equilateral triangle
of side length *s*.

What is the vertical component of the electrostatic force on the charge located on
vertex *P*?

^{kq2}/_{s2} sin 30°*kq* squared over *s* squared sine 30 degrees
^{kq2}/_{s2} cos 30°*kq* squared over *s* squared cosine 30 degrees
^{2kq2}/_{s2} sin 30°2*kq* squared over *s* squared sine 30 degrees
^{2kq2}/_{s2} cos 30°2*kq* squared over *s* squared cosine 30 degrees

Correct Response and Explanation (Show Correct ResponseHide Correct Response)

**D.** This question requires the examinee to describe an electric force
for a simple charge distribution. From Coulomb's law, the magnitude of the force on
*q* from the left-most charge on the base of the triangle is *F* =
^{kqq}/_{s2},*F* equals
fraction *kqq* over *s* squared, and the component
in the vertical direction is ^{kqq}/_{s2}
cos θ, where θ = 30°*kqq* over *s* squared
cosine theta, where theta equals 30 degrees. This force points in the positive *y*-direction.
The same applies for the force from the right-most charge. The net force is the
superposition of the two forces, which is *F* = ^{2kq2}/_{s2} cos 30°*F* equals >2*kq* squared over *s* squared cosine 30 degrees.

### Descriptive Statements:

- Demonstrate knowledge of the properties of magnets.
- Analyze the magnetic force on a moving charge in a magnetic field.
- Analyze the magnetic field for a current-carrying wire loop or solenoid.
- Apply Faraday's and Lenz's laws of induction to find the direction of an induced emf or current in a conducting loop.
- Describe the operation of devices such as electric motors, generators, and transformers.

### Sample Item:

A wire that carries conventional current *I* in the direction shown is in a magnetic field
that points into the plane of the page.

Which arrow shows the direction of the force on the wire?

Correct Response and Explanation (Show Correct ResponseHide Correct Response)

**C.** This question requires the examinee to analyze the magnetic force
on a moving charge in a magnetic field. The diagram shows a magnetic field that points
into the page. The wire carries conventional current, which implies positive charges
moving toward the top of the page. Since **F** = *I*ℓ
× **B** and ℓstart bold **F** end bold equals *I* samll l times start bold **B** end bold and small l represents
the direction of the conventional current *I*, the right-hand rule can be used to cross
**v** into **B**start bold **v** end bold into start bold **B** end bold. This shows that the force on the wire points to the left, as shown in response C.

### Descriptive Statements:

- Demonstrate knowledge of electromotive force, electric current, resistance, and Ohm's law.
- Describe characteristics of parallel and series circuits.
- Solve problems using Kirchhoff's laws for circuits.
- Analyze an electric circuit or component in terms of energy or power.

### Sample Item:

A resistor is connected in series across the terminals of an ideal battery of voltage
*V*. The resistor dissipates power *W* from the battery. A second,
identical resistor is connected in series with the first resistor. What is the total
power dissipated by the two resistors?

^{W}/_{2}*W* over 2
*W*
- 2
*W*
- 4
*W*

Correct Response and Explanation (Show Correct ResponseHide Correct Response)

**A.** This question requires the examinee to analyze an electric circuit
in terms of energy or power. The power dissipated by the resistor in the circuit is
given by *P* = *I**V*
= *W**P* equals *I**V*
equals *W*, where *I* is the current through
the resistor. When a second resistor is added to the circuit, the effective resistance
doubles. Since the voltage of the battery is the same, the current is
^{1}/_{2}*I*1 half *I*. Since there are
two identical resistors in the circuit, the voltage across each is
^{1}/_{2}*V*1 half *V* and the power dissipated
by each is *P* = ^{1}/_{4}*W**P* equals 1 fourth *W*.
Therefore, the total power dissipated is *P* = ^{1}/_{2}*W**P* equals 1 half *W*.