Home / AP® Exam / AP® Physics C: Electricity and Magnetism / Conservation of Electric Charges and Process of Charging MCQ
Three identical conducting spheres are mounted on insulating handles, as shown above. Spheres I and II have equal charges of \(+Q\) and are separated by a fixed distance. They repel each other with an electrostatic force of magnitude \(F\). Sphere III, initially uncharged, is first touched to sphere I, then to sphere II, and then removed. If the charge distribution on each sphere is assumed to always be spherical, the new magnitude of the electrostatic force between spheres I and II is
Correct Answer: \( \boxed{\mathrm{E}} \)
Initially,
Sphere I: \(+Q\), Sphere II: \(+Q\), Sphere III: \(0\).
Since the spheres are identical, when sphere III touches sphere I, the total charge \(Q\) is shared equally:
Sphere I \(=\dfrac{Q}{2}, \qquad \text{Sphere III}=\dfrac{Q}{2}.\)
Sphere III then touches sphere II, which still has charge \(Q\). The total charge is
\(Q+\dfrac{Q}{2}=\dfrac{3Q}{2},\)
so each sphere receives
\(\dfrac{3Q}{4}.\)
Final charges:
Sphere I \(=\dfrac{Q}{2}, \qquad \text{Sphere II}=\dfrac{3Q}{4}.\)
Initially,
\(F=k\dfrac{Q^{2}}{r^{2}}.\)
The new force is
\(F’= k\dfrac{\left(\dfrac{Q}{2}\right)\left(\dfrac{3Q}{4}\right)}{r^{2}} =\dfrac{3}{8} k\dfrac{Q^{2}}{r^{2}} =\dfrac{3F}{8}.\)
Therefore, the new electrostatic force between spheres I and II is
\( \boxed{\dfrac{3F}{8}}. \)
Therefore, the correct answer is (E).
When a positively charged rod is brought near, but does not touch, the initially neutral electroscope shown above, the leaves repel (I). When the electroscope is then touched with a finger, the leaves hang vertically (II). Next, when the finger and finally the rod are removed, the leaves repel again (III). During the process shown in Figure II,
Correct Answer: \( \boxed{\mathrm{B}} \)
Bringing the positively charged rod near the electroscope attracts electrons toward the knob, leaving the leaves positively charged so they repel each other.
In Figure II, when the electroscope is touched with a finger, it becomes temporarily grounded.
Because the nearby rod is positively charged, electrons are attracted from the ground (through the finger) into the electroscope.
The additional electrons neutralize the positive charge on the leaves, causing them to collapse and hang vertically.
After the finger is removed, the extra electrons remain on the electroscope. When the rod is finally removed, these excess electrons redistribute over the electroscope, leaving it negatively charged so the leaves repel once again.
Therefore, during Figure II, electrons flow from the finger into the electroscope.
Therefore, the correct answer is (B).
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
Correct Answer: \( \boxed{\mathrm{D}} \)
In Figure I, the negatively charged rod repels electrons from the knob into the leaves of the electroscope.
The leaves therefore acquire an excess of negative charge and repel each other, causing them to spread apart.
In Figure II, touching the electroscope with a finger grounds it. The excess electrons are repelled by the nearby negatively charged rod and flow from the electroscope into the ground, causing the leaves to collapse.
After the finger is removed while the rod is still nearby, the electroscope is left with a net positive charge. When the rod is finally removed, this positive charge redistributes uniformly over the electroscope.
Thus, in Figure III the leaves carry positive charge and once again repel each other.
Therefore, the leaves are negative in Figure I and positive in Figure III.
Therefore, the correct answer is (D).