Home / AP® Exam / AP® Physics C: Electricity and Magnetism / Redistribution of Charge between Conductors MCQ
Two identical spheres are \(10.0\,\mathrm{cm}\) apart and carry equal charges that create a force of \(4.00\times10^{-8}\,\mathrm{N}\) on each. Their diameters are much smaller than their separation distance. First one sphere is completely discharged. The spheres are then moved together until they touch, and finally they are moved to \(5.00\,\mathrm{cm}\) apart. The new force between the spheres is
Correct Answer: \( \boxed{\mathrm{C}} \)
Let the initial charge on each sphere be \(q\). The initial force is
\(F=k\dfrac{q^{2}}{r^{2}},\)
where \(r=10.0\,\mathrm{cm}\).
One sphere is then discharged, so the charges become \(q\) and \(0\). When the identical conducting spheres touch, the total charge \(q\) is shared equally, so each sphere has charge
\(\dfrac{q}{2}.\)
The spheres are then separated by \(5.00\,\mathrm{cm}\), which is half the original distance. The new force is
\(F’ = k\dfrac{\left(\frac{q}{2}\right)^2}{\left(\frac{r}{2}\right)^2} =k\dfrac{q^{2}/4}{r^{2}/4} =k\dfrac{q^{2}}{r^{2}} =F.\)
Therefore, the force remains unchanged:
\(F’=4.00\times10^{-8}\,\mathrm{N}.\)
Therefore, the correct answer is (C).
Two identical conducting spheres are charged to \(+2Q\) and \(-Q\), respectively, and are separated by a distance \(d\) (much greater than the radii of the spheres), as shown above. The magnitude of the force of attraction on the left sphere is \(F_{1}\). After the two spheres are made to touch and then are separated again by a distance \(d\), the magnitude of the force on the left sphere is \(F_{2}\). Which of the following relationships is correct?
Correct Answer: \( \boxed{\mathrm{E}} \)
Initially, the electrostatic force is given by Coulomb’s law:
\(F_{1}=k\dfrac{|(+2Q)(-Q)|}{d^{2}}=k\dfrac{2Q^{2}}{d^{2}}.\)
When the identical conducting spheres touch, the total charge is
\(+2Q-Q=+Q.\)
Since the spheres are identical, the total charge is shared equally, so each sphere has charge
\(\dfrac{Q}{2}.\)
After they are separated again by distance \(d\), the new force is
\(F_{2}=k\dfrac{\left(\frac{Q}{2}\right)^{2}}{d^{2}}=k\dfrac{Q^{2}}{4d^{2}}.\)
Therefore,
\(\dfrac{F_{1}}{F_{2}}=\dfrac{2}{1/4}=8.\)
Hence,
\(F_{1}=8F_{2}.\)
Therefore, the correct answer is (E).