Respuesta :
Answer:
[tex]\rho = 7.35\times \muC/m^3[/tex]
Explanation:
given,
Radius of the solid rod, R = 5.3 cm
Electric field strength,E = 22 kN/C
Let the volume charge density be ρ
From Gauss law
[tex]E = \dfrac{Rl}{2\epsilon_0}[/tex]
ε₀ is the permitivity of free space
R is the radius of the rod
and also,
[tex]\rho=\dfrac{2E\epsilon_0}{R}[/tex]
ρ is the volume charge density
[tex]\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}[/tex]
[tex]\rho = 7.35\times 10^{-6}\ C/m^3[/tex]
[tex]\rho = 7.35\times \muC/m^3[/tex]
Hence, the volume charge density is equal to [tex]\rho = 7.35\times \muC/m^3[/tex]
The volume charge density of the rod is equal to 7.35 [tex]\mu C/m^3[/tex]
Given the following data:
- Radius = 5.3 cm to m = 0.053 meter.
- Electric field strength = 22 kN/C.
Scientific data:
- Vacuum permittivity = [tex]8.85 \times 10^{-12}\;F/m[/tex]
How to calculate the volume charge density.
Assuming the length of the solid rod is long enough to approximate its electric field by using line symmetry, the flux through a length (L) of its surface can be equated with the charge enclosed in accordance with Gauss' law.
Mathematically, the volume charge density is given by this formula:
[tex]\rho = \frac{2E \epsilon_o}{R}[/tex]
Where:
- [tex]\rho[/tex] is the volume charge density.
- [tex]\epsilon_o[/tex] is the vacuum permittivity.
- E is the electric field strength.
- R is the radius.
Substituting the given parameters into the formula, we have;
[tex]\rho = \frac{2\times 22 \times 10^3 \times 8.85 \times 10^{-12}}{0.053}\\\\\rho = \frac{44000 \times 8.85 \times 10^{-12}}{0.053}\\\\\rho=7.35 \times 10^{-6}\;C/m^3[/tex]
Note: 1 [tex]\mu C[/tex] = [tex]1 \times 10^{-6}\;C[/tex]
Volume charge density = 7.35 [tex]\mu C/m^3[/tex]
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