Question

What is the electric field strength just outside the surface of a conducting sphere carrying surface charge density 1.4 μC/m2μC/m2?

Answer 1

Answer:

$E=158.19\frac{kN}{m}$

Explanation:

Gauss's theorem states that the flux of the electric field through a closed surface is equal to the the charge enclosed by the surface divided by the vacuum permittivity:

$\int\limits{\vec{E}\cdot \vec{dS}} \,=\frac{q}{\epsilon_0}$

The direction of the electric field ($\vec{E}$) just outside of a conductor is parallel to its surface ($\vec{S}$):

$\int\limits{\vec{E}\cdot \vec{dS}} \,=\int\limits{EdScos(0^\circ)} \,=E\int\limits{dS} \,=ES$

Recall that the surface charge density is defined as:

$\sigma=\frac{q}{S}$

Now, we get the electric field strength:

$ES=\frac{q}{\epsilon_0}\\E=\frac{q}{S\epsilon_0}\\E=\frac{\sigma}{\epsilon_0}\\E=\frac{1.4*10^{-6}\frac{C}{m^2}}{8.85*10^{-12\frac{C^2}{N\cdot m^2}}}\\\\E=158192.09\frac{N}{C}=158.19\frac{kN}{C}$