Question

A uniformly charged conducting sphere of diameter $3.5 \text{ cm}$ has a surface charge density of $20\mu\text{C m}^{-2}$. The total electric flux leaving the surface of the sphere is nearly}

• A. $57 \times 10^2 \text{ Vm}$
• B. $70 \times 10^2 \text{ Vm}$
• C. $87 \times 10^2 \text{ Vm}$
• D. $35 \times 10^3 \text{ Vm}$

Hint:

Flux depends only on the total enclosed charge, not the distribution. $\Phi = \sigma A / \varepsilon_0$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Total electric flux leaving a closed surface depends only on the net charge enclosed within it, according to Gauss's Law.
First, calculate the total charge on the spherical surface using the charge density.
Step 2: Key Formula or Approach:
Total charge $q = \sigma \cdot A$, where $A = 4\pi r^2$ is the surface area.
Gauss's Law for electric flux: $\Phi_E = \frac{q}{\varepsilon_0}$.
Step 3: Detailed Explanation:
Given diameter $D = 3.5 \text{ cm} \implies$ radius $r = 1.75 \text{ cm} = 1.75 \times 10^{-2} \text{ m}$.
Surface charge density $\sigma = 20 \mu\text{C m}^{-2} = 20 \times 10^{-6} \text{ C m}^{-2}$.
Calculate surface area $A$: \[ A = 4\pi r^2 = 4 \times 3.14159 \times (1.75 \times 10^{-2})^2 \] \[ A \approx 12.566 \times 3.0625 \times 10^{-4} \approx 38.48 \times 10^{-4} \text{ m}^2 \] Calculate total charge $q$: \[ q = \sigma \times A = (20 \times 10^{-6}) \times (38.48 \times 10^{-4}) \] \[ q = 769.6 \times 10^{-10} \text{ C} \approx 7.696 \times 10^{-8} \text{ C} \] Calculate electric flux $\Phi_E$: \[ \Phi_E = \frac{q}{\varepsilon_0} = \frac{7.696 \times 10^{-8}}{8.85 \times 10^{-12}} \] \[ \Phi_E \approx 0.8696 \times 10^4 = 8696 \text{ V}\cdot\text{m} \] This value can be approximated to $8700$, which is $87 \times 10^2$.
Note: The options incorrectly use the unit Weber (Wb), which is for magnetic flux. The numerical value is what is being tested.
Step 4: Final Answer:
The numerical value is $87 \times 10^2$.