Question

A uniformly charged conducting sphere of $2.4\,\text{m}$ diameter has a surface charge density of $80.0\,\mu\text{C/m}^{2}$.
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?

Hint:

Use R = diameter/2 = 1.2 m. Charge Q = sigma times 4 pi R^2, then total flux = Q / epsilon_0.

Answer 1

Step 1: Find the radius from the diameter. Diameter \(= 2.4\,\text{m}\), so:
\[R = \frac{2.4}{2} = 1.2\,\text{m}\]
Step 2 (part a): The charge equals the surface charge density times the surface area of the sphere:
\[Q = \sigma\times A = \sigma\times 4\pi R^2\]
Step 3: Substitute \(\sigma = 80.0\,\mu\text{C m}^{-2} = 80.0\times10^{-6}\,\text{C m}^{-2}\) and \(R = 1.2\,\text{m}\):
\[Q = (80.0\times10^{-6})\times 4\pi\,(1.2)^2\]
Step 4: Do the arithmetic:
\[Q = (80.0\times10^{-6})\times(18.10) = 1.45\times10^{-3}\,\text{C}\]
Step 5 (part b): The total flux leaving the surface follows from Gauss's law:
\[\Phi = \frac{Q}{\epsilon_0} = \frac{1.45\times10^{-3}}{8.85\times10^{-12}}\]
Step 6: Evaluate:
\[\Phi = 1.64\times10^{8}\,\text{N m}^2\,\text{C}^{-1}\]
\[\boxed{Q = 1.45\times10^{-3}\,\text{C},\quad \Phi = 1.64\times10^{8}\,\text{N m}^2\,\text{C}^{-1}}\]

Answer 2

Surface-charge-density to flux directly.
Step 1: Combine the two relations \(Q = \sigma\,(4\pi R^2)\) and \(\Phi = Q/\epsilon_0\) into a single expression for the flux:
\[\Phi = \frac{\sigma\,(4\pi R^2)}{\epsilon_0}\]
Step 2: With \(R = 1.2\,\text{m}\), compute the area:
\[4\pi R^2 = 4\pi(1.2)^2 = 18.10\,\text{m}^2\]
Step 3: First get the charge:
\[Q = \sigma\times 18.10 = (80.0\times10^{-6})(18.10) = 1.45\times10^{-3}\,\text{C}\]
Step 4: Now the flux:
\[\Phi = \frac{(80.0\times10^{-6})(18.10)}{8.85\times10^{-12}} = 1.64\times10^{8}\,\text{N m}^2\,\text{C}^{-1}\]
\[\boxed{Q = 1.45\times10^{-3}\,\text{C},\quad \Phi = 1.64\times10^{8}\,\text{N m}^2\,\text{C}^{-1}}\]