Free Sample AP Physics C: Electricity and Magnetism

When the switch S is opened, the equivalent resistance, , of the parallel circuit increases. This results in the increase of the of the entire circuit. When the of the entire circuit increases, the total current, I, must decrease. This is because the total potential difference, V, is constant. The decrease in total I results in a decrease in the V across the 54 Ω resistor. Since the total V has not changed, while the V across the 54 Ω has decreased, the V across the 10 Ω resistor must increase. This results in an increase in the current through the 10 Ω resistor. The current is not doubled, though, because that would require the V across the 10 Ω resistor to have doubled. This would only have occurred if the V across the 54 Ω resistor had halved. That would have required the circuit’s current to have halved as well. This would have occurred only if the 10 Ω had a much greater resistance.

If the mass of the +q charge is m, then its acceleration is

Because a should decrease as r increases, you can eliminate () and (). The graph in () best depicts an inverse square relationship between a and r.

First, consider the whole circuit: the battery is the same, so the total voltage is unchanged. The equivalent resistance increases because we removed a parallel branch. By V = IR, the total current must decrease. Now apply Ohm’s law to resistor X only: The reduced total current all runs through resistor X, but resistor X can’t change its own resistance. So by V = IR, the voltage across bulb X is smaller than before. Kirchoff’s loop rule says that the voltage across X and Y must sum to V. . . so with smaller voltage across X, there must be greater voltage across Y. Now apply Ohm’s law to bulb Y: greater voltage with unchanged resistance means greater current, too.

Ampere’s law says that the line integral of the magnetic field around a closed line is . How much current is enclosed by the circumference of the wire? The areal current density j times the area enclosed. The cross-sectional area enclosed here is just the area of the circular wire, .

The potential due to a point charge is , where d is the distance from the charge. Both objects produce the same potential. So the equation with the bigger charge in the numerator must also have a bigger distance in the denominator and by the same factor because the equation has no squares or square roots.

The electric field at the center of the ring is zero because the field caused by any charge element is canceled by the field due to the charge on the other side of the ring. The electric field decreases as by Coulomb’s law, so a long distance away from the ring the field goes to zero. The field is nonzero near the ring, though, because each charge element creates a field pointing away from the ring, resulting in a field always along the axis.

Because the plate separation d is very small compared to the radius of the cylinder, it is reasonable to approximate the capacitor as a parallel plate capacitor with plate separation d and plate area 2πrl. (This makes sense: Because d << r, the electric field is relatively constant between the plates and is perpendicular to the plates.) Therefore, the approximate capacitance is

Even without making this observation, a few choices can be eliminated. Compared to the formula , choice () has incorrect units, so it can be eliminated. Choice () does not include the length of the capacitor. However, we know from the simpler case of a parallel plate capacitor that the dimension of the plates is relevant to the value of the capacitance, so we can safely eliminate choice ().

The force on +Q is equal in magnitude and opposite in direction to the force on −Q, so the net force is zero. However, both forces produce a clockwise torque about the center of the dipole, so the net torque is clockwise.

I is false: The direction of the force depends on the sign of the charge on the particle, so it cannot be known for any charged particle. II is true: The magnitude of the electric force is simply |F| = QE. III is true: The rate of change in voltage with respect to position in the direction parallel to the field is simply E, with potential decreasing in the direction of the field. IV is true: The rate of change in potential with respect to position in the direction perpendicular to the field is always equal to zero.

The charges within the bar are moving in a magnetic field, so they feel a magnetic force. Using the conventional-current description, positive charges feel an upward force, causing positive charge to accumulate at the top end of the bar (leaving negative charge on the bottom end of the bar). This charge continues to accumulate until the electric force created by the charge imbalance cancels the magnetic force, mathematically qE = qvB. The magnitude of the potential difference between points A and B is then |V | = |−∫E · dr| = |vBd|. The electric field points down (producing a force that opposes the magnetic force), and recalling that electric fields point from high to low potential, we see that point A is at the higher potential.