Question

Two metal spheres of radii \(r_{1}=10.0 \mathrm{~cm}\) and \(r_{2}=20.0 \mathrm{~cm},\) respectively, have been positively charged so that each has a total charge of 100. \(\mu\) C. a) What is the ratio of their surface charge distributions? b) If the two spheres are connected by a copper wire, how much charge flows through the wire before the system reaches equilibrium?

Short answer

The ratio of their surface charge distributions is 4:1, and 50 μC of charge flows through the wire to reach equilibrium.

Step-by-step solution

Calculate surface charge density for each sphere

To calculate the surface charge density (\(\sigma\)), we divide the charge (\(Q\)) by the surface area of each sphere: $$ \sigma = \frac{Q}{4\pi r^2} $$ For sphere 1: $$ \sigma_1 = \frac{100\ \mu\C}{4\pi (10.0 \mathrm{~cm})^2} $$ For sphere 2: $$ \sigma_2 = \frac{100\ \mu\C}{4\pi (20.0 \mathrm{~cm})^2} $$ (Note: In the equations above, we can keep the unit of charge as \(\mu\)C and the unit of radius as cm because we are ultimately looking for the ratios, and the units will cancel out.)

Find the ratio of surface charge densities

Now, let's divide \(\sigma_1\) by \(\sigma_2\) to find the ratio of surface charge densities: $$ \frac{\sigma_1}{\sigma_2} = \frac{\frac{100\ \mu \mathrm{C}}{4\pi (10.0 \mathrm{~cm})^2}}{\frac{100\ \mu \mathrm{C}}{4\pi (20.0 \mathrm{~cm})^2}} $$ Simplifying the expressions, we get: $$ \frac{\sigma_1}{\sigma_2} = \frac{(20.0 \mathrm{~cm})^2}{(10.0 \mathrm{~cm})^2} = 4 $$ So, the ratio of their surface charge distributions is 4:1.

Calculate the charge flow between the spheres

When the spheres are connected with a copper wire, charges will flow between them until their potentials become equal. Potentials of spheres 1 (\(V_1\)) and 2 (\(V_2\)) are given by: $$ V_1 = \frac{kQ}{r_{1}} $$ $$ V_2 = \frac{kQ}{r_{2}} $$ where \(k\) is the electrostatic constant (\(k \approx 9 \times 10^9 \,\mathrm{Nm^2/C^2}\)), \(Q\) is the charge on the spheres, and \(r_1\) and \(r_2\) are the radii. At equilibrium, the potentials will be equal, so let's set \(V_1 = V_2\): $$ \frac{kQ}{r_{1}} = \frac{kQ}{r_{2}} $$ Rearranging the equation to solve for the equilibrium charge \(Q_{eq}\), we get: $$ Q_{eq} = \frac{r_{1}+r_{2}}{r_{2}} \cdot Q $$ Substitute the sphere radii and initial charge, and we find that \(Q_{eq} \approx 150\ \mu \mathrm{C}\). Now that we know the equilibrium charge, we can determine the charge flow between the spheres. For sphere 1, the charge flow is given by: $$ \Delta Q_1 = Q_{eq} - Q = 150\ \mu \mathrm{C} - 100\ \mu \mathrm{C} = 50 \ \mu \mathrm{C} $$ This means that 50 \(\mu\)C of charge flows through the wire to sphere 1, and therefore the same amount of charge, 50 \(\mu\)C, flows from sphere 2.