Question

A long pipe of outer radius ????1=3.50 cm and inner radius ????2=2.98 cm carries a uniform charge density of 9.22 mC/m3. Assuming that the pipe is sufficiently long to consider it infinitely long, use Gauss's law to calculate the electric field ???? at a distance ????=7.35 cm from the centerline of the pipe. Use ????0=8.85×10−12 C/N·m2 for the the permittivity of free space.

Answer 1

Answer:

2388078.86544 N/C

Explanation:

$\rho$ = Charge density = 9.22 mC/m³

r = Distance = 7.35 cm

$r_o$ = Outer radius = 3.5 cm

$r_i$ = Inner radius = 2.98 cm

l = Length of cylinder

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

V = Volume

E = Electric field

Charge is given by

$Q=\rho V\\\Rightarrow Q=\rho\pi l(r_o^2-r_i^2)$

Area

$A=2\pi rl$

From Gauss law the flux through a cylindrical surface is given by

$EA=\frac{Q}{\epsilon_0}\\\Rightarrow E=\frac{Q}{\epsilon_0A}\\\Rightarrow E=\frac{\rho\pi l(r_o^2-r_i^2)}{\epsilon_02\pi rl}\\\Rightarrow E=\frac{\rho(r_o^2-r_i^2)}{\epsilon_02r}\\\Rightarrow E=\frac{9.22\times 10^{-3}(0.035^2-0.0298^2)}{8.85\times 10^{-12}\times 2\times 0.0735}\\\Rightarrow E=2388078.86544\ N/C$

The electric at the given distance is 2388078.86544 N/C