Question

A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) carrying 65.0 uC of charge is placed such that its center is 40.0 cm away from the center of the first sphere (A). a. If the two conducting spheres are now connected by a thin conducting wire, what are the new charges on the spheres? b. Now, the conducting wire is cut and the spheres are released from rest. What are their speeds when they are far apart (infinite distance away) from each other? Take the mass of each conductor to be 80.0 grams. Ignore gravity. Assume that the potential is zero at infinity.

Answer 1

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

$Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C$

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

$\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J$

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

$\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}$

hence, the speed of the spheres is 37.45 m/s