Question

a very long nonconducting cylinder of radius ro and length L (ro < L) passes a uniform charge of density. determine the electric field outside of cylinder

Answer 1

Answer on Question#53294 - Physics - Field Theory

A very long nonconducting cylinder of radius $\rho$ and length $L$ ( $\rho < L$ ) passes a uniform charge of density $\alpha$ . Determine the electric field outside of cylinder.

Solution:

According to the Gauss's law (considering that the cylinder is very long) the electric flux through the very long coaxial cylindrical surface is equal to the inner charge divided by $\varepsilon_0$ . Let's consider a piece of such cylinder with radius $R$ and length $l$ . Since the surface area of this piece is $A = 2\pi Rl$ , the electric flux through this surface is

$$\Phi = A \cdot E (R) = 2 \pi R l \cdot E (R),$$

where $E(R)$ – is the electric field created by the charged cylinder. Electrical charge surrounded by this surface (if $R > \rho$ ) is $Q = 2\pi \rho l \cdot \alpha$ . Therefore we obtain

$$\Phi = \frac {Q}{\varepsilon_ {0}}$$ $$2 \pi R l \cdot E (R) = \frac {2 \pi \rho l \cdot \alpha}{\varepsilon_ {0}}$$ $$E (R) = \frac {\rho}{R} \frac {\alpha}{\varepsilon_ {0}}$$

For $R < \rho$ there are no inner electrical charge, thus

$$E (R) = 0, \qquad R < \rho$$

Answer:

$$E (R) = \left\{ \begin{array}{l l} 0, & R < \rho \\ \frac {\rho}{R} \frac {\alpha}{\varepsilon_ {0}}, & R > \rho \end{array} \right.$$

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