Question

A sphere of radius 5 cm has a charge of 31.41 \(\mu\)C. Calculate the surface density of charge.

Hint:

In physics problems, look for numbers that are multiples or fractions of common constants like \(\pi\) or \(e\). Here, \(31.41\) is clearly intended to be related to \(10\pi\), which simplifies the calculation significantly by allowing \(\pi\) to be cancelled out. Always convert all given quantities to their S.I. units before starting calculations.

Answer 1

Step 1: Understanding the Concept:
Surface charge density (\(\sigma\)) is a measure of the amount of electric charge distributed over a given surface area. It is defined as the charge per unit area.
Step 2: Key Formula or Approach:
The formula for surface charge density is:
\[ \sigma = \frac{Q}{A} \] where:
- \(\sigma\) is the surface charge density.
- \(Q\) is the total charge on the surface.
- \(A\) is the total surface area.
For a sphere, the surface area is given by \(A = 4\pi r^2\), where \(r\) is the radius.
Step 3: Detailed Explanation:
First, we need to list the given values and convert them to S.I. units.
- Charge, \(Q = 31.41 \, \mu\text{C} = 31.41 \times 10^{-6} \, \text{C}\).
- Radius, \(r = 5 \, \text{cm} = 0.05 \, \text{m}\).
We can notice that \(31.41 \approx 10 \times 3.141 \approx 10\pi\). This approximation can simplify the calculation. So let's use \(Q \approx 10\pi \times 10^{-6}\) C.
Next, calculate the surface area (\(A\)) of the sphere:
\[ A = 4\pi r^2 \] \[ A = 4\pi (0.05 \, \text{m})^2 \] \[ A = 4\pi (0.0025 \, \text{m}^2) \] \[ A = 0.01\pi \, \text{m}^2 \] Now, calculate the surface charge density (\(\sigma\)):
\[ \sigma = \frac{Q}{A} \] \[ \sigma = \frac{10\pi \times 10^{-6} \, \text{C}}{0.01\pi \, \text{m}^2} \] The \(\pi\) terms cancel out.
\[ \sigma = \frac{10 \times 10^{-6}}{0.01} \, \frac{\text{C}}{\text{m}^2} \] \[ \sigma = \frac{10 \times 10^{-6}}{10^{-2}} \, \frac{\text{C}}{\text{m}^2} \] \[ \sigma = 10 \times 10^{-6} \times 10^2 \, \frac{\text{C}}{\text{m}^2} \] \[ \sigma = 10 \times 10^{-4} \, \frac{\text{C}}{\text{m}^2} \] \[ \sigma = 1.0 \times 10^{-3} \, \frac{\text{C}}{\text{m}^2} \] Step 4: Final Answer:
The surface density of charge on the sphere is \(1.0 \times 10^{-3}\) C/m\(^2\).