APC Unit 3 Problem Set Mistakes

Suppose that an electron (charge -e) could orbit a proton (charge +e) in a circular orbit of constant radius R. Assuming that the proton is stationary and only electrostatic forces act on the particles, which of the following represents the kinetic energy of the two-particle system?

(1/8πε₀)e²/R Notes: the system is in UCM so Fc = mv²/r = ke²/r and KE = mv²/r. Solve for v² and plug it into KE formula.

A closed surface, in the shape of a cube of side a, is oriented as shown above in a region where there is a constant electric field of magnitude E parallel to the x-axis. The total electric flux through the cubical surface is

0 Notes: No net charge is enclosed

Three 6-microfarad capacitors are connected in series with a 6-volt battery. The energy stored in EACH capacitor is

12 Notes: U = CV²/2 where C = 6 and V = 2

A 12-volt storage battery, with an internal resistance of 2Ω, is being charged by a current of 2 amperes. Under these circumstances, a voltmeter connected across the terminals of the battery will read

16V Notes: voltage drop across internal resistance is 4V, and added to given voltage of battery is 16V

when E = 0, what does that mean about voltage?

constant

Two objects on a horizontal frictionless surface each have charge +Q and each are fixed in place on the x axis at the same distance d from the origin as shown in the figure above. A particle of charge -q constrained to move along the y axis is released from rest. After release, the particle will

exhibit oscillatory motion

Two conducting spheres, X and Y. have the same positive charge +Q, but different radii (rx > ry) as shown above. The spheres are separated so that the distance between them is large compared with either radius. If a wire is connected between them, in which direction will current be directed in the wire?

from Y to X Notes: need to calculate using kq/r

The nonconducting hollow sphere of radius R shown above carries a large charge +Q, which is uniformly distributed on its surface. There is a small hole in the sphere. A small charge +q is initially located at point P. a distance r from the center of the sphere. If k = 1/4πεo, what is the work that must be done by an external agent in moving the charge +q from P through the hole to the center O of the sphere?

kqQ(1/R - 1/r) Notes: W = q∆V = q(VR - Vr) where VR = kQ/R and Vr = kQ/r

When a negatively charged rod is brought near, but does not touch, the initially uncharged electroscope shown above, the leaves spring apart (I). When the electroscope is then touched with a finger, the leaves collapse (II). When next the finger and finally the rod are removed, the leaves spring apart a second time (III). The charge on the leaves is

negative in I, positive in III Notes: Charge separation in I, Charging by induction in III (opposite charge of the rod)

The point charge Q shown above is at the center of a metal box that is isolated, ungrounded, and uncharged. Which of the following is true?

net charge on box is Q other answers that were WRONG: 1) The potential inside the box is zero, 2) the electric field inside the box is constant, 3) the electric field outside the box is zero everywhere, 4) the electric field outside the box is the same as if only the point charge (and not the box) were there Notes: The charge +Q induces a charge -Q on the inner surface of the box, inducing a charge +Q on the outer surface

Two positive charges of magnitude q are each a distance d from the origin A of a coordinate system, as shown above. Where is the electric field least in magnitude?

right in the middle of the charges Notes: while the magnitude of the electric forces would be double, they would cancel out because they go in opposite directions

Points R and S are each the same distance d from two unequal charges, +Q and +2Q, as shown above. The work required to move a charge -Q from point R to point S is

setup answer: 0 Notes: V = ΣkQ/r, with the symmetry VR = VS. W = q∆V = 0

Two positive charges of magnitude q are each a distance d from the origin A of a coordinate system as shown. At which of the following points is the electric potential greatest in magnitude?

setup: answer: A Notes: V = ΣkQ/r and since both charges are positive, the largest potential is at the closest point to the two charges (it is more mathematically complex than that, but this reasoning works for the choices given)

A physics problem starts: "A solid sphere has charge distributed uniformly throughout. . . " It may be correctly concluded that the

sphere is note made of metal Notes: For charge to be distributed throughout an object, it must not be a conductor, otherwise the charge would move to the surface of the object

A positive charge +Q located at the origin produces an electric field Eo at point P (x = +1, y = 0). A negative charge -2Q is placed at such a point as to produce a net field of zero at point P. The second charge will be placed on the

x-axis where x < 0 Notes: placement is -√2 NOT +√2 because E field due to +Q points to right. If second charge were placed at √2 then +Q would be attracted to -2Q and point to the right again, making the field stronger. That's why we have to place the second charge at -√2 to cancel out

The potential of an isolated conducting sphere of radius R is given as a function of the charge q on the sphere by the equation V = kq/R. If the sphere is initially uncharged, the work W required to gradually increase the total charge on the sphere from zero to Q is given by which of the following expressions?

∫(0 → Q) kQ/r dq Notes: The incremental amount of work required to bring a small amount of charge is dW = V(dq) where V is the potential relative to infinity at that time, which is kq/R (q being the amount of charge currently on the sphere)

A helium nucleus (charge +2q and mass 4m) and a lithium nucleus (charge +3q and mass 7m) are accelerated through the same electric potential difference, V0 . What is the ratio of their resultant kinetic energies, Klithium/KHelium?

3/2 Notes: V = ∆U/q → U = qV → 3qV/2qV

How much energy is dissipated by the 1.5-ohm resistor with a current of 2A running through it in 60 seconds?

360 J Notes: Energy = Pt = I²Rt

Two identical conducting spheres are charged to +2Q and -Q. respectively, and are separated by a distance d. The magnitude of the force of attraction on the left sphere is F1. After the two spheres are made to touch and then are reseparated by distance d the magnitude of the force on the left sphere is F2. Which of the following relationships is correct?

Before touching, this product is 2Q². After touching and separating, the new charges are (+2Q - Q)/2 = Q/2. The new force is ¼Q²k/2. Therefore, F₁ = 8F₂

An air gap capacitor originally has capacitance C. If a thin sheet of metal is placed halfway between the plates of the capacitor without touching either plate, the effective capacitance is

C Notes: with the metal sheet, there are now 2 capacitors in series with d/2 → capacitance = 2C → Ceq = [1/2C + 1/2C]⁻¹

A point charge is placed at the center of an uncharged, spherical, conducting shell of radius R. The electric fields inside and outside the sphere are measured. The point charge is then moved off center a distance R/2 and the fields are measured again. What is the effect on the electric fields?

Changed inside but not changed outside Notes: The arrangement of the field lines inside will change as the charge is moved as it is due to that charge. Outside the sphere, the induced surface charge is evenly distributed around the outer surface.

Two charged particles, each with a charge of +q, are located along the x-axis at x = 2 and x = 4, as shown above. Which of the following shows the graph of the magnitude of the electric field along the x-axis from the origin to x = 6?

E = 0 at the midpoint and is large near each of the charges, growing to infinity as the charge is approached

A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radius r with r < R is used to calculate the magnitude of the electric field E at a distance r from the center of the sphere. Which of the following equations results from a correct application of Gauss's law for this situation?

E(4πr²) = Qr³/ε₀R³ Notes: Left side: EA = E(4πr²) (area of the Gaussian surface) Right side: Q enclosed = fraction of Q inside = Q × (volume of Gaussian surface/volume of sphere)

The electric field of two long coaxial cylinders is represented by lines of force as shown above. The charge on the inner cylinder is +Q. The charge on the outer cylinder is

Here's the set up of the problem: Answer: -3Q Notes: If four lines is proportional to +Q, then 8 inward pointing lines is proportional to -2Q. This would be the net charge enclosed, so to get -2Q in total we would need -3Q + Q

A uniform spherical charge distribution has radius R.. Which of the following is true of the electric field strength due to this charge distribution at a distance r from the center of the charge?

It is directly proportional to r when r < R. Notes: Linear (proportional to r) inside, proportional to 1/r2 outside

Positive charge Q is uniformly distributed over a thin ring of radius a that lies in a plane perpendicular to the x-axis. with its center at the origin 0, as shown above. What graph best represents the electric field along the positive x-axis?

Note: At the center of the ring, the field is zero due to symmetry/cancellation.

The figure above shows a spherical distribution of charge of radius R and constant charge density ρ. Which of the following graphs best represents the electric field strength E as a function of the distance r from the center of the sphere?

Notes: Linear (proportional to r) inside, proportional to 1/r2 outside

Concentric conducting spheres of radii a and 2a bear equal but opposite charges +Q and -Q. respectively. Which of the following graphs best represents the electric potential V as a function of r ?

Notes: Outside the spheres E = 0 so V is constant. Once inside the negative shell, the potential is that for a positively charged conducting sphere (constant inside, proportional to 1/r outside)

Two infinite parallel sheets of charge perpendicular to the x-axis have equal and opposite charge densities as shown above. The sheet that intersects x = -a has uniform positive surface charge density; the sheet that intersects x = +a has uniform negative surface charge density. Which graph best represents the plot of electric field as a function of x ?

Notes: The field from an infinite sheet of charge is uniform and, in this case, equal in magnitude (will be the same across at each point b/w the sheets), pointing away from the sheet on the left and toward the sheet on the right. The E field cancels outside the sheets.

A battery or batteries connected to two parallel plates produce the equipotential lines between the plates (-2, -1, 0, 1, 2 V). Which of the following configurations is most likely to produce these equipotential lines?

Notes: The potential difference between the plates is 4V, this can be produced with two 2V batteries in series (note the positive plate is on the left)

Two small spheres having charges of +2Q and -Q are located 12 centimeters apart. The potential of points lying on a line joining the charges is best represented as a function of the distance x from the positive charge by which of the following?

Notes: V = ΣkQ/r, positive and approaching infinity as it nears the positive charge and negative and approaching negative infinity near the negative charge. Since the positive charge is larger, the zero point is closer to the smaller charge.

Two particles, each with charge -Q, are fixed a distance L apart. Each particle experiences net electric force F. A particle with charge +q is now fixed midway between the original two particles. As a result, the net electric force experienced by each negatively charged particle is reduced to F/2. What is q?

Q/8 Notes: 1) F = kQ²/L² 2) F/2 = kQq/(L/2)² = kQ²/2L² solving for q in number 2 gets us Q/8

In the circuit shown, the emf's of the batteries are given, as well as the currents in the outside branches and the resistance in the middle branch. What is the magnitude of the potential difference between X and Y ?

Setup: Answer: 12V Notes: Kirchhoff's junction rule applied at point X gives 2 A = I + 1 A, so the current in the middle wire is 1 A. Summing the potential differences through the middle wire from X to Y gives - 10 V - (1A)(2 Ω) = -12 V

In the circuit shown above, what is the resistance R ?

Setup: Answer: 4 Notes: The current through R is found using the junction rule at the top junction, where 1 A + 2 A enter giving I = 3 A. Now utilize Kirchhoff's loop rule through the left or right loops: (left side) + 16 V - (1 A)(4 Ω) - (3 A)R = 0 giving R = 4 Ω

If the ammeter in the circuit above reads zero, what is the resistance R?

Setup: Answer: 6 Notes: For the ammeter to read zero means the junctions at the ends of the ammeter have the same potential. For this to be true, the potential drops across the 1 Ω and the 2 Ω resistor must be equal, which means the current through the 1 Ω resistor must be twice that of the 2 Ω resistor. This means the resistance of the upper branch (1 Ω and 3 Ω) must be 1⁄2 that of the lower branch (2 Ω and R) giving 1 Ω + 3 Ω = 1⁄2 (2 Ω + R)

The graph above shows the electric potential V in a region of space as a function of position along the x-axis. At which point would a charged particle experience the force of greatest magnitude?

Setup: Answer: D Notes: F = qE, E = -dV/dx, F is largest when E is largest and E is largest where voltage has the greatest slope

When the switch S is open in the circuit shown above, the reading on the ammeter A is 2.0 A. When the switch is closed, the reading on the ammeter is

Setup: Answer: increased slightly but not doubled Notes:

The resistance R is

Setup: Answer: Notes: Utilizing Kirchhoff's loop rule starting at the upper left and moving clockwise: - (2 A)(0.3 Ω) + 12 V - 6 V - (2 A)(0.2 Ω) - (2A)(R) - (2A)(1.5 Ω) = 0

At which point does the electric field have the greatest magnitude?

Setup: Answer: B Notes: E has the greatest magnitude where the lines are closest

In this region the electric potential relative to infinity is proportional to

Setup: Answer: Q₁/r + Q₂/r₂ Notes: Relative to infinity, on the outer surface of the larger shell, the potential is k(Q₁ + Q₂)/r₂. Once inside there is no more change to the potential due to Q₂, but still varies as 1/r due to Q₁ until the final position is reached

A long time after the switch has been closed, the current supplied by the battery is

Setup: Answer: V/(R₁ + R₂) Notes: When the capacitor is fully charged, the branch with the capacitor is "closed" to current, effectively removing it from the circuit for current analysis.

Immediately after the switch is closed, the current supplied by the battery is

Setup: Answer: V/R₁ Notes: When the switch is closed, the circuit behaves as if the capacitor were just a wire, shorting out the resistor on the right.