Physics C—Electricity and Magnetism Practice Exam—Multiple-
Choice Solutions
1 . B —An electric field exists regardless of the amount of charge placed in it, and regardless of
whether any charge at all is placed in it. So both experimenters must measure the same field (though
they will measure different forces on their test charges).
2 . D —You could use Gauss’s law to show that the field outside the sphere has to decrease as 1/r 2 ,
eliminating choices B and E. But it’s easier just to remember that an important result of Gauss’s law
is that the electric field inside a conductor is always zero everywhere , so D is the only possibility.
3 . B —While in the region between the plates, the negatively charged electron is attracted to the
positive plate, so bends upward. But after leaving the plates, there is no more force acting on the
electron. Thus, the electron continues in motion in a straight line by Newton’s first law.
4 . C —This is a Newton’s third law problem! The force of A on B is equal (and opposite) to the force
of B on A . Or, we can use Coulomb’s law: The field due to A is k (2Q )/(4 m)^2 . The force on B is
QE = k 2 QQ /(4 m)^2 . We can do the same analysis finding the field due to B and the force on A to get
the same result.5 . A —The charge is in equilibrium, so the horizontal component of the tension must equal the electric
force. This horizontal tension is 0.1 N times sin 30° (not cosine because 30° was measured from the
vertical ), or 0.05 N. The electric force is qE , where q is 0.002 C. So the electric field is 0.050
N/0.002 C. Reduce the expression by moving the decimal to get 50/2, or 25 N/C.
6 . D —The answer could, in principle, be found using the integral form of Coulomb’s law. But you
can’t do that on a one-minute multiple-choice problem. The electric field will point down the page—
the field due to a positive charge points away from the charge, and there’s an equal amount of charge
producing a rightward field as a leftward field, so horizontal fields cancel. So, is the answer B or
D? Choice B is not correct because electric fields add as vectors. Only the vertical component of the
field due to each little charge element contributes to the net electric field, so the net field must be
less than kQ /R 2 .
7 . A —Use the symmetry of the situation to see the answer. Because the infinitely large plane looks
the same on the up side as the down side, its electric field must look the same, too—the field must
point away from the plane and have the same value.
8 . C —Another way to look at this question is, “Where would a small positive charge end up if
released near these charges?” because positive charges seek the smallest potential. The positive
charge would be repelled by the +2Q charge and attracted to the −Q charges, so would end up at
point C . Or, know that potential due to a point charge is kq /r . Point C is closest to both −Q charges,
so the r terms will be smallest, and the negative contribution to the potential will be largest; point C
is farthest from the +2Q charge, so the r term will be large, and the positive contribution to the
potential will be smallest.
9 . A —A positive charge is forced from high to low potential, which is generally to the left; and the