Chapter 23: Q56P (page 683)

The electric field in a particular space is $\vec{E} = (x + 2) \hat{i} N / C$ , with xin meters. Consider a cylindrical Gaussian surface of radius that is coaxial with the x-axis. One end of the cylinder is at $x = 0$ . (a) What is the magnitude of the electric flux through the other end of the cylinder at $X = 2.0 m$ ? (b) What net charge is enclosed within the cylinder?

Expert verified

a) Magnitude of the electric flux through the other end of the cylinder at x = 2.0 m is $0.50 N . m^{2} / C$

b) Net charge enclosed by the cylinder is $2.2 x 10^{- 12} C$ .

Step by step solution

$\vec{E} = (x + 2) \hat{i} N / C$

cylindrical Gaussian surface of radius 20 cm

Using the statement of Gauss law determine the charge enclosed by the cylinder

There is no flux through the sides, so we have two contributions to the flux, one

from the x = 2.0 m end

$ϕ_{2} = + (2.0 + 2) N / C (π {(0.20 m)}^{2}) = 0.50 N . m^{2} / C$

and one from the x = 0 end

$ϕ_{2} = + (2 N / C) (π {(0.20 m)}^{2}) = 0.25 N . m^{2} / C$

Magnitude of the electric flux through the other end of the cylinder at x = 2.0 m is $0.50 N . m^{2} / C$

By Gauss’ law, we have

$q_{enc} = ε_{0} (ϕ_{2} + ϕ_{0}) = (8.85 \times 10^{- 12} C^{2} / N . m^{2}) \times (0.25 N . m^{2} / C) = 2.2 \times 10^{- 12} C$

The net charge enclosed by the cylinder is $2.2 \times 10^{- 12} C$

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