How much charge is on a conducting sphere of diameter $20 \mu \mathrm{m}$ if its surface potential is $50 \mathrm{mV}$ ?

Question

Quizplus · Accepted Answer

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

How Much Charge Is on a Conducting Sphere of Diameter 20μm20 \mu \mathrm{m}20μm

How Much Charge Is on a Conducting Sphere of Diameter $20 \mu \mathrm{m}$