Answer :
Answer:
The electric field at the cylinder surface = 80.19 kN/C
Electric flux via the cylinder [tex]\mathbf{\phi_E = 828.63 Nm^2/C}[/tex]
Explanation:
Given that:
For the filament
The length = 4.95 m
The charge = 2.00 µC
The charge per unit length for the filament can be computed as:
[tex]\lambda = \dfrac{q}{l}[/tex]
[tex]\lambda = \dfrac{2}{4.5}\mu C/m[/tex]
Using Gauss's law:
[tex]\phi_E = \oint E^{\to}*dA^{\to}[/tex]---- (1)
where;
electric flux = [tex]\phi_E[/tex]
permittivity of free space = [tex]\varepsilon_o[/tex]
electric field = E
surface area = dA
However, the electric flux [tex]\phi_E[/tex] via the cylinder can be expressed as:
[tex]\phi_E = \dfrac{\lambda l'}{\varepsilon_o}[/tex]
Equation (1) can now be rewritten as:
[tex]\dfrac{\lambda l'}{\varepsilon_o} = \oint E^{\to}*d(2 \pi rl')[/tex]
[tex]|E| =(\dfrac{\lambda l'}{\varepsilon_o 2 \pi rl' })[/tex]
replacing the values into the above equation:
[tex]|E| =(\dfrac{(\dfrac{2}{4.5} \mu C/m) (1.65 \ cm)}{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (10 \ cm )(1.65 cm) })[/tex]
[tex]|E| =(\dfrac{(\dfrac{2}{4.5} \times 10^{-6} C/m) }{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (0.1 \ m ) })[/tex]
[tex]\mathbf{|E| =80.19 \ kN/C}[/tex]
Thus, the electric field at the cylinder surface = 80.19 kN/C
The electric flux now is calculated using the said formula:
[tex]\phi_E = \dfrac{\lambda l'}{\varepsilon_o}[/tex]
[tex]\phi_E = \dfrac{(\dfrac{2}{4.5}\times 10^{-6} \ C)(0.0165 \ m)}{8.85 \times 10^{-12} \ C^2/Nm^2}[/tex]
[tex]\mathbf{\phi_E = 828.63 Nm^2/C}[/tex]