How much charge is on a conducting sphere of diameter 20 μm if its surface potential is 50 mV?

Question

Quizplus · Accepted Answer

The charge $Q$ on a conducting sphere can be calculated using the formula $Q = 4\pi\epsilon_0RV$, where $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$), $R$ is the radius of the sphere, and $V$ is the potential. Given the diameter is $20 \, \mu\text{m}$, the radius $R = 10 \, \mu\text{m} = 10 \times 10^{-6} \, \text{m}$, and the potential $V = 50 \, \text{mV} = 50 \times 10^{-3} \, \text{V}$. Substituting these values into the formula gives $Q = 4\pi(8.85 \times 10^{-12})(10 \times 10^{-6})(50 \times 10^{-3})$, which simplifies to approximately $56 \times 10^{-18} \, \text{C}$, or $56 \times 10^{-18}$ Coulombs.

How Much Charge Is on a Conducting Sphere of Diameter