Question

A non conducting rod with radius equal to 5cm has a volume charge density of (p) 1.6x10^-5 C/m^3 .

What is the electric flux passing through a cylindrical plane of distance:
a) 4.15 cm
b) 6.50 cm
From the outside of the rod?

Answer 1

Answer on Question #73248 - Physics / Electromagnetism

A non-conducting rod with radius equal to $R = 5$ cm has a volume charge density of $\rho = 1.6 \times 10^{-5} \, \text{C/m}^3$. What is the electric flux passing through a cylindrical plane of distance

a) $4.15\,\text{cm}$

b) $6.50\,\text{cm}$

from the outside of the rod?

Solution:

The Gauss law for the electric flux through a closed surface $S$ enclosing any volume $V$ gives

$$\Phi = \frac{Q}{\varepsilon_0},$$ $$Q = \int \rho \, dV$$

where $Q$ is a total charge enclosed within volume $V$, $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$ is an electric constant. Thus

a)

$$\Phi = \frac{\int \rho \, dV}{\varepsilon_0} = \frac{\rho \pi R^2 h}{\varepsilon_0}$$

The electric flux per unit length of a cylindrical plane

$$\Phi = \frac{\rho \pi R^2}{\varepsilon_0} = \frac{1.6 \times 10^{-5} \times 3.14 \times 0.05^2}{8.85 \times 10^{-12}} = 1.42 \times 10^4 \, \text{V}$$

b) In this case answer would be the same, because total charge enclosed within volume not changed.

Answers:

a) $\Phi = 1.42 \times 10^4 \, \text{V}$

b) $\Phi = 1.42 \times 10^4 \, \text{V}$

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