After connecting, the charges equalize.

Repulsive force is kq^2/r^2 = (9E9)(12E-6)^2/(.6)^2

A) 5.4 N, attractive

B) zero

C) 3.6 N, repulsive

D) 3.6 N, attractive

E) 5.4 N, repulsive

F1 = kq^2/r^2

F2 = k(q/2)^2/r^2

A) 1/4 F

B) 1/16 F

C) 1/8 F

D) 1/2 F

Newton’s Third Law: For every action there is an equal and opposite reaction

A) F

B) 3F

C) F/sqrt(3)

D) sqrt(3)F

E = lambda / (2*pi*8.85E-12*r)

So, to double E, we need to cut r in half

A) D/2

B) 2D

C) sqrt(2)D

D) D/4

E) D/sqrt(2)

A) is always perpendicular to the surface of the conductor

B) is perpendicular to the surface of the conductor only if it’s a sphere, a cylinder, or flat sheet

C) can have nonzero components perpendicular to and parallel to the surface

D) is always parallel to the surface

E) is always zero because the electric field is zero inside conductors

Since there’s no charge on the cone, total flux = 0 (flux on bottom and sloped sides sum to 0).

phi(bottom) + phi(curved) = 0

-(4550)(pi*.0211)^2 + phi(curved) = 0

A) 6.36 Nm^2/C

B) 10.4 Nm^2/C

C) 1.24 Nm^2/C

D) 25.5 Nm^2/C

E) 82.1 Nm^2/C

A) No work is required to move the negative charge from point A to point B

B) Work is required to move the negative charge

C) The work done on the charge depends on the distance between A and B

D) The negative charge performs work in moving from point A to point B

E) Work is done in moving the negative charge from point A to point B

Don’t mix this up witha conducting sphere where the potential is zero inside the sphere

A) The potential at the center of the sphere is the same as the potential at the surface.

B) The potential at the surface is higher than its center

C) The potential at the center is the same as the potential at infinity

D) The potential at the center of the sphere is zero.

E) The potential is highest at the center of the sphere

phi(entire sphere) = q(enclosed)/8.85E-12

Phi(entire sphere) / 2 gives us the answer

A)14

B) 11

C) 17

D) 20

E) 23

Ex = -dv/dx = (-2y – 2x)

Ey = -dv/dy = (-2y – 1)

Set both equal to 0 and solve for x and y.

B) x = 1 m, y = 1 m

C) x = .5 m, y = .5 m

D) x = .5 m, y = 1 m

E) x = 0 m, y = 0 m

phi = q(enc)/8.85E-12 = 15 +(-40) /8.85E-12

A) + 2.8

B) -1.1

C) 1.1

D) -2.8

E) -.47

d = 3i + 4j

Va – Vb = E.d = (4i + 3j)(3i +4j) = 24

B) 27

C) 30

D) 24

E) 11

E is conservative (unless created by magnetic fields, in which case it wouldn’t be constant), so its line integral on any clsoed loop is always 0.

A) E(2*pi*r)

B) E*pi*r^2

C) 0

D) -E(2*pi*r)

A) charge

B) energy

C) electric field

D) voltage

E) electric force

Va = k(4E-6)/.3 + k(-6E-6)/.5

Vb = k(-6E-6)/.3 + k(4E-6)/.5

Va-Vb

A) 96 V

B) 0 V

C) 48 V

D) 48 kV

E) 96 kV

More current goes through the path with bulb A as there is less total resistance in that path. More current = more brightness. Since the current does not separate between B and C, they have the same brightness

A) B and C have equal brightness, and A is the dimmest

B) All three bulbs have the same brightness

C) A is the brightest, C is dimmest, B is in between

D) A and B have equal brightness, and C is the dimmest

E) A is the brightest, B and C are equal but not as bright as A

A) Charge on C2 decreases

B) The charge on C2 increases

C) the charge on C2 either inreases or decreases depending on the value of the dielectric constant

D) The charge on C2 remains the same

E) The charge on C2 either increases or decreases depending on whetehr C2 is greater or less than C1

The centripetal force is given by m*v(perpendicular)^2/r, where v(perpendicular) = vsin(theta)

Thus, r = mvsin(theta)/(qB)

Use loop rule to set up that Epsilon – i1r1 – i2r2 = 0

A) -3 V

B) -10 V

C) -1 V

D) +3 V

E) +19 V

P = v^2/R

(R ends up being 121 ohms but 120 is the closest value given)

A) 8.0 milli-ohms

B) 100 ohms

C) 6.0 milli-ohms

D) 240 ohms

E) 120 ohms

*A cap is discharged in 5 time constants, thus

5(RC) = 50E-3 s… solve for C to get 1E-5 F –> 10E-6 F

A) 100E-6 F

B) 2E-6 F

C) 10E-6 F

D) 20E-6 F

E) 200E-6 F

E = vB for no deflection, but also E = V/d (where v = velocity and V = potential difference).

vB = V/d –> V = vBd

A) 94 V

B) 190 V

C) 1900 V

D) 3800 V

E) 600 V

A) at the speed of sound in the metal

B) at the speed of light

C) at the speed of sound in air

D) at none of these speeds

E) at nearly the speed of light

Capacitors act the opposite as resistors for equivalence

A) 1.7E-6 F

B) 8.0E-6 F

C) 12E-6 F

D) 4.0E-6 F

E) 7.1E-6 F

A) toward the east

B) toward the west

C) toward the south

D) toward the north

E) downward

Magnetic forces never do work. ya dingus.

A) An essential piece of information is missing

B) e(v x B)

C) zero

D) F.d

E) Fdcos(theta)

A) the end to end potential difference

B) the electron drift velocity

C) the electric field

D) the current

E) the current density

Frequency is indpendent of velocity in the cone

A) f/2

B) f/4

C) 4f

D) f

E) 2f

Maximizes the equivalent capacitance, so more charge can be stored by both capacitors and hence stored energy is maximized

A) in series across the source

B) in parallel across the source

C) it doesn’t matter because the stored energy is the same either way.

Just memorize it I guess

r^2 so you only need to multiply by 2 to get 4

A) 1/4 the diameter of the original wire

B) 2 times the diameter of the original wire

C) 4 times the diameter of the original wire

D) The same diameter

E) 1/2 the diameter of the original wire

Uo = Q^2/2Co

New C = (8.85E-12)(A)/(2d)

–> U = Q^2/2C = Q^2/(2(Co/2)) = Q^2/Co = 2Uo

A) Uo

B) 2Uo

C) 4Uo

D) Uo/2

E) Uo/4

The changing B field produces an EMF which opposes the battery.

|EMF| = dphi(B)/d(t) = A dB/dt

V(battery) – |EMF| = iR

4 – .12^2(dB/dt) = .1(10)

dB/dt = 208 T/s

Must be decreasing to set direction of EMF correct by Lens’ law

A) 420 T/s, increasing

B) 208 T/s, decreasing

C) 140 T/s, increasing

D) 140 T/s, decreasing

E) 208 T/s, increasing

Since the B field is in the plane of the loop, the flux of B over the loop is ALWAYS 0. No change in flux = no change in induced emf.

A) from b to a

B) from a to b

C) No induced current through the resistor

|EMF| = N(dphi/dt) = N(BA – 0)/dt

A) .032 V

B) .0032 V

C) 0 V

D) .32 V

E) 1.0 V

Integral of B.dl = Mu(o)*I = Mu(o)*2

A) 2.5E-7

B) none of these

C) 2.5E-6

D) 3.8E-6

E) 3.0E-7

|EMF| = N (dphi/dt) = NA (dB/dt).

Doubling both N and dB/dt multiplies the whole thing by 4

A) Epsilon/4

B) Epsilon

C) 2*Epsilon

D) Epsilon/2

E) 4*Epsilon

Energy Density = B^2/(2*Mu(o))

Where B = Mu(o)*i/(2*pi*r)

A) 3.7E-5 J/m^3

B) 1.2E-4 J/m^3

C) 3.6E-4 J/m^3

D) 7.3E-5 J/m^3

E) The density cannot be determined without knowing the voluem

C is the only possibility that doesn’t cause a change in the fluxx across the turns of the solenoid

A) Rotate the coil about the y-axis

B) Rotate about the x-axis

C) Rotate about the z-axis

D) Move the coil toward point P

E) Change the current in the solenoid

Remember: i = I(o)*e^(-t/T), where T = L/R

A) .234 s

B) 14.0 ms

C) 2.73 s

D) 1.00 s

E) 2.190 s

B = -i*B(loop) – k*B(wire)

(Directions determined by right hand rule)

B(loop) = Mu(o)*10/(2*pi*.5)

B(wire) = (Mu(o)/2)*(q(.5)^2/(.5^2+.5^2)^3/2

(Picture not shown)

A) zero

B) -8E-6 T k

C) -4E-6 T i – 8E-6 T k

D) -4E-6 T i – 4E-6 T k

E) +4E-6 T i – 4.0E-6 T k

Just memorize it.

A) (1/2)pi*k*R^4

B) pi*k*R^3

C) (2/3)pi*k*R^3

D) r*k*R

Epsilon = L (di/dt)

1000 = (100E-3)(di/dt)

A) 100 A/s

B) 10 A/s

C) 10,000 A/s

D) 1 A/s

E) 1000 A/s

The loop sees a magnetic field pointed into the paper, decreasing. To counteract this change in flux, the loop creates an emf in the clockwise direction (b to a)

A) from a to b

B) from b to a

C) there is no induced current through the resistor

D) the induced current will be oscillating back and forth

The B fields shown are increasing, applying lens’ law gives the answer

A) conterclockwise in both coils

B) clockwise in both coils

C) counterclockwise in the east coil and clockwise in the west coil

D) clockwise in the east coil and counterclockwise in the west coil

Loop 2 experiences an increasing magnetic field pointed to the left, applying lenz’ law forces current to move clockwise

A) counter-clockwise

B) clockwise

C) no current will be induced because velocity is constant

D) no current will be induced because of Lenz’s law

T = Mu*B*Sin(theta)

T = (IA)*B*sin(theta)

1.1 = 4A(.6)sin(60)

A = .53 m^2

Note: Theta is the perpendicular – the plane of the loop’s angle given = 60 degrees.

A) .53 m^2

B) .20 m^2

C) .80 m^2

D) .4 m^2

E) .26 m^2

V = IX where X = 2*pi*f*L

Solve for L

A) 1.43 µH

B) 1.36 µH

C) 9.20 µH

D) 4.42 µH

E) 1.97 µH

Vs = Vp (Ns/Np)

A) 458

B) 240

C) 110

D) 22

E) 4

Is = Ip(Np/Ns)

^different than Voltage transformations

A) 700

B) 356

C) 114

D) 14

E) 4

Just memorize it.

A) 4 f0

B) 2 f0

C) f0

D) f0/2

E) f0/4

Remember lab

A) The impedance has its maximum value.

B) The reactance of the inductor is zero.

C) The reactance of the capacitor is zero.

D) The reactance due to the inductor and capacitor has its maximum value.

E) The current amplitude is a maximum.

A) It decreases.

B) It increases.

C) It does not change.

D) It increases or decreases depending on the sign of the phase angle.

A) a resistor

B) an inductor

C) a capacitor

D) an inductor or a capacitor

v (velocity)

k (spring constant)

m (mass)

q

i

1/C

L

I = V/Xc

Calculated f may be in kHz, but the answer is in MHz for some reason.

A) 32.6 MHz

B) 5.20 MHz

C) 32.6 kHz

D) 5.20 kHz

E) 32.6 Hz

(b) 700 volts

(a) Does the source voltage lag or lead the current?

(b) What is the voltage amplitude of the source?

Z = sqrt(R^2 + (Xl – Xc)^2)

A) 98 V

B) 96 V

C) 93 V

D) 91 V

E) 88 V

Xc = 1/(2*pi*f*C)

Xl = 2*pi*L*f

Xc = 2(Xl) and solve for f

A) 14 Hz

B) 13 Hz

C) 16 Hz

D) 17 Hz

E) 19 Hz

U = .5*L*i^2

*remember that the given current is in rms and must be converted into the peak current by multiplying by sqrt(2) to get 1.838 A

A) 0.32 J

B) 0.16 J

C) 0.48 J

D) 0.64 J

E) 0.80 J

tan(phase angle) = (Xl – Xc)/R

A) 11.8 mH

B) 124 mH

C) 212 mH

D) 61.9 mH

E) 19.7 mH

where phi is the phase angle

Just use P(avg) = I^2rms*R and then multiply by cos(39)

A) 930 kW

B) 47 kW

C) 93 kW

D) 190 kW

B = (Mu(o)*i)/(2*pi*r) where i = (epsilon.o)(area)(change in E field/change in time)

i = (epsilon.o)(area)(change in E field/change in time)

A) parallel to one another and perpendicular to the direction of wave propagation.

B) parallel to one another and parallel to the direction of wave propagation.

C) perpendicular to one another and perpendicular to the direction of wave propagation.

D) perpendicular to one another and parallel to the direction of wave propagation.

|E x B|.

A) xy-plane.

B) +z-direction.

C) -z-direction.

D) -x-direction.

E) -y-direction.

A) equally divided between the electric and magnetic fields.

B) mostly in the electric field.

C) mostly in the magnetic field.

D) all in the electric field.

E) all in the magnetic field.

A) 4F

B) 2F

C) F

D) F/2

E) F/4

A) its speed, wavelength, and frequency all decrease.

B) its speed and wavelength decrease, but its frequency stays the same.

C) its speed decreases but its wavelength and frequency both increase.

D) its speed decreases but its frequency and wavelength stay the same.

E) its speed increases, its wavelength decreases, and its frequency stays the same.

I = Io*cos(theta)^2

if Io is initially unpolarized, I = Io/2.

A) 53°

B) 73°

C) 85°

D) 80°

The angles should be exactly 90° from either of the other polarizers

A) 40° and 90°

B) 40° and 130°

C) 40° and 140°

D) 90° and 130°

E) 90° and 140°

A) infinity

B) 2f

C) between f and 2f

D) f

E) between the lens and f

m = -i/p

*since the image is bigger, the location of the image must be negative

(1/30) + (1/-45) = 2/r

A) 40 cm

B) 60 cm

C) 100 cm

D) 140 cm

E) 180 cm

The image is in a negative orientation because it’s formed behind the mirror (virtual)

A) 2.2 cm in front of the mirror

B) 4.4 cm in front of the mirror

C) 9.0 cm in front of the mirror

D) 1.0 cm in front of the mirror

E) 20 cm in front of the mirror

A) 1.5 mm

B) 3.0 mm

C) 4.5 mm

D) 6.0 mm

E) 9.0 mm