Question 53: Electrostats & Capacitance

Question

A uniformly charged conducting sphere of 0.2 m diameter has a surface charge density of $70 \mu \mathrm{Cm}^{-2}$. The electric flux leaving the surface of the sphere is:

Accepted Answer

To find the flux through the sphere, apply GaussÃ¢ÂÂs law:
$$\Phi = \frac{Q}{\varepsilon_0}$$
where $Q$ is the total charge on the sphere and $\varepsilon_0 = 8.85	imes10^{-12}\,\mathrm{C^2/(N\cdot m^2)}$.
Compute the radius:
$$R = \frac{d}{2} = \frac{0.2\,\mathrm m}{2} = 0.1\,\mathrm m.$$
Find the total charge $Q$:
$$Q = \sigma 	imes 4\pi R^2       = 70	imes10^{-6}\,\frac{\mathrm C}{\mathrm m^2}\,	imes 4\pi\,(0.1\,\mathrm m)^2      = 2.8\pi	imes10^{-6}\,\mathrm C      \approx 8.80	imes10^{-6}\,\mathrm C.$$
Apply GaussÃ¢ÂÂs law:
$$\Phi = \frac{Q}{\varepsilon_0}      = \frac{2.8\pi	imes10^{-6}}{8.85	imes10^{-12}}      \approx 9.94	imes10^5\;\mathrm{N\cdot m^2/C}.$$
So the flux is about $9.9	imes10^5\;\mathrm{N\cdot m^2/C}$, which matches Option A.

Answer

$9.9 	imes 10^5 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

Answer

$9.9 	imes 10^6 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

Answer

$8.9 	imes 10^5 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

Answer

$8.9 	imes 10^6 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

Electrostats & Capacitance

A uniformly charged conducting sphere of 0.2 m diameter has a surface charge density of 70 \mu \mathrm{Cm}^{-2}. The electric flux leaving the surface of the sphere is:

Choose the correct answer: