Explain briefly.
(c) If the two spheres are connected by a metal wire, will charge flow from the big sphere to the little
sphere, or from the little sphere to the big sphere? Explain briefly.
(d) Which of the following two statements is correct? Explain briefly.
i. If the two spheres are connected by a metal wire, charge will stop flowing when the electric
field at the surface of each sphere is the same.
ii. If the two spheres are connected by a metal wire, charge will stop flowing when the electric
potential at the surface of each sphere is the same.
(e) Explain how the correct statement you chose from part (d) is consistent with your answer to (c).Solutions to Practice Problems
1 . C— Electric field is a vector . Look at the field at the center due to each charge. The field due to the
left-hand charge points away from the positive charge; i.e., to the right; the field due to the right-hand
charge points to the left. Because the charges are equal and are the same distance from the center point,
the fields due to each charge have equal magnitudes. So the electric field vectors cancel! E = 0.
2 . D— Electric potential is a scalar . Look at the potential at the center due to each charge: Each charge
is distance a /2 from the center point, so the potential due to each is kQ /(a /2), which works out to 2kQ
/a . The potentials due to both charges are positive, so add these potentials to get 4kQ /a .
3 . B— If the potential difference between plates is, say, 100 V, then we could say that one plate is at
+100 V and the other is at zero V. So, the potential must change at points in between the plates. The
electric field is uniform and equal to V /d (d is the distance between plates). Thus, the potential
increases linearly between the plates, and A must have twice the potential as B.
4 . A— The electric field by definition is uniform between parallel plates. This means the field must be
the same everywhere inside the plates.
5 . C— We have cylindrical symmetry, so use Gauss’s law. Consider a Gaussian surface drawn within the
outer shell. Inside a conducting shell, the electric field must be zero, as discussed in the chapter. By
Gauss’s law, this means the Gaussian surface we drew must enclose zero net charge. Because the inner
cylinder carries charge +Q , the inside surface of the shell must carry charge −Q to get zero net charge
enclosed by the Gaussian surface. What happens to the −2Q that is left over on the conducting shell? It
goes to the outer surface.6 . (a) Like charges repel, so the charges are more likely to spread out from each other as far as
possible.
(b) “Conducting spheres” mean that the charges are free to move anywhere within or onto the surface
of the spheres. But because the charges try to get as far away from each other as possi-ble, the
charge will end up on the surface of the spheres. This is actually a property of conductors—charge
will always reside on the surface of the conductor, not inside.
(c) Charge will flow from the smaller sphere to the larger sphere. Following the reasoning from parts
(a) and (b), the charges try to get as far away from each other as possible. Because both spheres
initially carry the same charge, the charge is more concentrated on the smaller sphere; so the
charge will flow to the bigger sphere to spread out. (The explanation that negative charge flows
from low to high potential, and that potential is less negative at the surface of the bigger sphere, is
also acceptable here.)