The value of the x-component of the electric field is -475213.968 newtons per coulomb.
Procedure - Determination of the magnitude of an electric field at a given point
In this question we shall apply Gauss' Law to determine the magnitude of the electric field ([tex]E_{x}[/tex]), in newtons per coulomb, rapidly and based on the assumptions of uniform charge distribution and cylindrical symmetry.
[tex]\frac{Q_{enc}}{\epsilon_{o}} = \oint\,\vec E\,\bullet d\vec A[/tex] (1)
Where:
- [tex]Q_{enc}[/tex] - Enclosed charge, in coulombs.
- [tex]\epsilon_{o}[/tex] - Vacuum permitivity, in quartic second-square amperes per kilogram-cubic meter.
- [tex]\vec E[/tex] - Electric field vector, in newtons per coulomb.
- [tex]\vec A[/tex] - Area vector, in square meters.
Based on all assumptions, we simplify (1) as follows:
[tex]\frac{\lambda\cdot l}{\epsilon_{o}} = E \cdot (2\pi\cdot r\cdot l)[/tex]
And the equation of the x-component of the electric field is:
[tex]E = \frac{\lambda}{2\pi\cdot \epsilon_{o}\cdot r}[/tex] (2)
Where [tex]\lambda[/tex] is the linear charge density, in coulomb per meter.
If we know that [tex]\lambda = -2.3\times 10^{-6}\,\frac{C}{m}[/tex] and [tex]a = 0.087\,m[/tex], then the electric field produced by the line of charge at point P is:
[tex]E = \frac{\left(-2.3\times 10^{-6}\,\frac{C}{m} \right)}{2\pi\cdot \left(8.854\times 10^{-12}\,\frac{s^{4}\cdot A^{2}}{kg\cdot m^{3}} \right)\cdot (0.087\,m)}[/tex]
[tex]E_{x} = -475213.968 \,\frac{N}{C}[/tex]
The value of the x-component of the electric field is -475213.968 newtons per coulomb. [tex]\blacksquare[/tex]
Remark
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