Physics, 01.02.2021 23:40 agpraga23ovv65c

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?

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Answer from: TanelleK9546

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 ( $E_{x}$ ), in newtons per coulomb, rapidly and based on the assumptions of uniform charge distribution and cylindrical symmetry.

$\frac{Q_{enc}}{\epsilon_{o}} = \oint\,\vec E\,\bullet d\vec A$ (1)

Where:

$Q_{enc}$ - Enclosed charge, in coulombs. $\epsilon_{o}$ - Vacuum permitivity, in quartic second-square amperes per kilogram-cubic meter. $\vec E$ - Electric field vector, in newtons per coulomb. $\vec A$ - Area vector, in square meters.

Based on all assumptions, we simplify (1) as follows:

$\frac{\lambda\cdot l}{\epsilon_{o}} = E \cdot (2\pi\cdot r\cdot l)$

And the equation of the x-component of the electric field is:

$E = \frac{\lambda}{2\pi\cdot \epsilon_{o}\cdot r}$ (2)

Where $\lambda$ is the linear charge density, in coulomb per meter.

If we know that $\lambda = -2.3\times 10^{-6}\,\frac{C}{m}$ and $a = 0.087\,m$ , then the electric field produced by the line of charge at point P is:

$E = \frac{\left(-2.3\times 10^{-6}\,\frac{C}{m} \right)}{2\pi\cdot \left(8.854\times 10^{-12}\,\frac{s^{4}\cdot A^{2}}{kg\cdot m^{3}} \right)\cdot (0.087\,m)}$

$E_{x} = -475213.968 \,\frac{N}{C}$

The value of the x-component of the electric field is -475213.968 newtons per coulomb. $\blacksquare$

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What is Ex(P), the value of the x-component of the electric field produced by by the line of charge

Answer from: abbyramirez52

The answer is below

Explanation:

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:

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

Answer from: Quest

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Answer from: Quest

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