What is Ex(P), the value of the x-component of the electric field produced by by the line of charge at point P which is located at (x,y) = (a,0), where a = 8.7 cm?

We are going to use Gauss’ law to find the electric field equation. Since electric field is coming from an infinite line of charge, hence it is going out in a radial direction.

Therefore we use the area of the electric field which passes through, forming a Gaussian cylinder. We neglect the ends of the area.

Hence:

[tex]\int\limits {E} \, dA=\frac{Q_{enc}}{\epsilon_o}\\\\E(2\pi rL)= \frac{\lambda L}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r\epsilon_o} \\\\Given \ that:\\\\r=a=8.7\ cm=0.087\ m, \lambda=-2.3 \mu C/cm=-2.3*10^{-4}\ C/m,\epsilon_o=8.85*10^{-12}F/m.\\\\Hence:\\\\E=\frac{-2.3*10^{-4}}{2\pi *0.087*8.85*10^{-12}}=-4.75*10^7\ N/C[/tex]

xero099 xero099 · Answer 2 · 2022-01-18T13:51:22+01:00

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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