Part A = +z

Part B = -z

Part C = +y

Part D = at a –45° angle in the xz plane

The electric and magnetic field vectors at a specific point in space and time are illustrated. Based on this information, in what direction does the electromagnetic wave propagate?

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Part A = 75000 m/s

Part B = 75000

A charged particle travels undeflected through perpendicular electric and magnetic fields whose magnitudes are 3000 N/C and 40 mT, respectively.

Find the speed of the particle if it is a proton.

Find the speed of the particle if it is an alpha particle. (An alpha particle is a helium nucleus-a positive ion with a double positive charge of +2e.)

Part A = 5 N/C

Part B = -x̂

Part C = -x̂

A metal cube with sides of length a = 1 cm is moving at velocity v = 1 m/s ŷ across a uniform magnetic field B = 5 T ẑ. The cube is oriented so that four of its edges are parallel to its direction of motion (i.e., the normal vectors of two faces are parallel to the direction of motion).

Find E, the magnitude of the electric field inside the cube.

What is the direction of the induced electric field?

If one replaces the conducting cube with one that has positive charge carriers, what is the direction of the induced electric field?

Part A = 6.31 * 10^{-9} C

Part B = 3.79 * 10^{-8} J

Part B = 2700 V/m

A 12.0 V battery remains connected to a parallel plate capacitor with a plate area of 0.264 m^{2} and a plate separation of 4.44 mm.

What is the charge on the capacitor?

How much energy is stored in the capacitor?

What is the electric field between its plates?

Part A = U = Q^{2} / (2C)

Part B = U = (CV^{2}) / 2

Part C = U/U_{0} = 0.5

Part D = U/U_{0} = 2

Part E = U = (ε_{0}AV^{2}) / 2d

Part F = U = (ε_{0}AdE^{2}) / 2

Part G = u = (ε_{0}E^{2}) / 2

Find the energy U of the capacitor in terms of C and Q by using the definition of capacitance and the formula for the energy in a capacitor.

Find the energy U of the capacitor in terms of C and V by using the definition of capacitance and the formula for the energy in a capacitor.

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U_{0}. The capacitor remains connected to the battery while the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U_{0}.

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U_{0}. The capacitor is then disconnected from the battery and the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U_{0}.

A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem introduction to obtain the formula for the energy U of the capacitor.

A parallel-plate capacitor has area A and plate separation d, and it is charged so that the electric field inside is E. Use the formulas from the problem introduction to find the energy U of the capacitor.

Find the energy density u of the electric field in a parallel-plate capacitor. The magnitude of the electric field inside the capacitor is E.

Part A = 0 J

Part B = 1 J

Part C = greater than the magnitude of the electric field at point B.

What is the work W_{AB} done by the electric force to move a 1 C charge from A to B?

What is the work W_{AD} done by the electric force to move a 1 C charge from A to D?

The magnitude of the electric field at point C is