Question 2.3: Find the electric field a distance z above one end of a stra...

Introduction to Electrodynamics – Solution Manuals [EXP-2863]

Find the electric field a distance z above one end of a straight line segment of length L (Fig. 2.7) that carries a uniform line charge λ. Check that your formula is consistent with what you would expect for the case $z \gg L$ .

Question Data is a breakdown of the data given in the question above.

Distance above one end of the straight line segment: z
Length of the straight line segment: L
Uniform line charge: λ

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The given expression represents the electric field, denoted as E, at a point in space due to a line of charge with linear charge density λ. The line of charge extends from 0 to L along the x-axis.

Step 1:

To calculate the electric field, we need to integrate the contribution from each infinitesimally small charge element along the line. We start by considering the electric field component in the z-direction, denoted as Ez.

Step 2:

Define the variables and parameters Let z be the distance from the line of charge to the point where we want to calculate the electric field. Let x be the distance along the line of charge. Let η^2 = z^2 + x^2. Let θ be the angle between the line of charge and the line connecting the charge element to the point where we want to calculate the electric field.

Step 3:

Determine the electric field component in the z-direction The electric field component in the z-direction can be calculated using the equation: Ez = (1 / (4πε₀)) ∫ (λ dx / η^2) cos θ, where the integral is taken from 0 to L.

Step 3:

Simplify the expression Using the definitions of η^2 and cos θ, we can simplify the expression for Ez as: Ez = (1 / (4πε₀)) ∫ (λ dx / η^2) cos θ = (1 / (4πε₀)) ∫ (λ dx / η^2) (z / η) = (1 / (4πε₀)) ∫ (λ dx / η^3) z

Step 4:

Evaluate the integral Integrating the expression with respect to x from 0 to L, we get: Ez = (1 / (4πε₀)) λ z ∫ (1 / η^3) dx, where the integral is taken from 0 to L.

Step 5:

Evaluate the integral limits Evaluating the integral limits, we have: Ez = (1 / (4πε₀)) λ z [(1 / z^2) (x / η)] evaluated from 0 to L = (1 / (4πε₀)) λ z [(1 / z^2) (L / η) - (1 / z^2) (0 / η)] = (1 / (4πε₀)) λ z [(1 / z^2) (L / η)] = (1 / (4πε₀)) λ/ z (L / (z^2 + L^2)^(1/2))

Step 6:

Simplify the expression Simplifying the expression, we have: Ez = (1 / (4πε₀)) (λ / z) (L / (z^2 + L^2)^(1/2))

Step 7:

Similarly, we can calculate the electric field component in the x-direction, denoted as Ex, using a similar process.

Step 8:

Determine the electric field component in the x-direction The electric field component in the x-direction can be calculated using the equation: Ex = -(1 / (4πε₀)) ∫ (λ dx / η^2) sin θ, where the integral is taken from 0 to L.

Step 9:

Simplify the expression Using the definitions of η^2 and sin θ, we can simplify the expression for Ex as: Ex = -(1 / (4πε₀)) ∫ (λ dx / η^2) (x / η) = -(1 / (4πε₀)) ∫ (λx dx / η^3)

Step 10:

Evaluate the integral Integrating the expression with respect to x from 0 to L, we get: Ex = -(1 / (4πε₀)) λ ∫ (x / η^3) dx, where the integral is taken from 0 to L.

Step 5:

Evaluate the integral limits Evaluating the integral limits, we have: Ex = -(1 / (4πε₀)) λ [(1 / η^2) (-1 / η)] evaluated from 0 to L = -(1 / (4πε₀)) λ [(1 / η^2) (-1 / η) - (1 / η^2) (-1 / η)] = -(1 / (4πε₀)) λ [(1 / z) - (1 / (z^2 + L^2)^(1/2))]

Step 6:

Simplify the expression Simplifying the expression, we have: Ex = -(1 / (4πε₀)) λ [(1 / z) - (1 / (z^2 + L^2)^(1/2))]

Step 7:

Finally, we can express the electric field E as a vector in terms of its components Ex and Ez. E = (1 / (4πε₀)) (λ / z) [(-1 + (z / (z^2 + L^2)^(1/2))) x + (L / (z^2 + L^2)^(1/2)) z]

This is the theoretical explanation of the given expression for the electric field. It shows the step-by-step process of calculating the electric field components and combining them to obtain the final expression.

Final Answer

E_z=\frac{1}{4\pi \varepsilon _0} \int_{0}^{L}\frac{\lambda dx}{\eta ^2}\cos \theta ;\left(\eta ^2=z^2+x^{2};\cos \theta =\frac{z}{\eta } \right)

= $\frac{1}{4\pi \varepsilon _0} \lambda z\int_{0}^{L}{}\frac{1}{\left(z^{2}+x^{2} \right)^{3/2} }dx$

= $\frac{1}{4\pi \varepsilon _0} \lambda z\left[\frac{1}{z^2}\frac{x}{\sqrt{z^2+x^{2} } } \right] \mid ^{L}_{0}=\frac{1}{4\pi \varepsilon _0}\frac{\lambda }{Z}\frac{L}{\sqrt{z^2+L^2} }$ .

E_x=-\frac{1}{4\pi \varepsilon _0} \int_{0}^{L}{} \frac{\lambda dx}{\eta ^2}\sin \theta =-\frac{1}{4\pi \varepsilon _0}\lambda \int{} \frac{xdx}{\left(x^{2}+z^2 \right)^{3/2} }

= $-\frac{1}{4\pi \varepsilon _0}\lambda \left[-\frac{1}{\sqrt{x^{2}+z^2 } } \right] \mid ^{L}_{0}=-\frac{1}{4\pi \varepsilon _0} \lambda \left[\frac{1}{z}-\frac{1}{\sqrt{z^2+L^2} } \right]$ .

$E =\frac{1}{4\pi \varepsilon _0} \frac{\lambda }{z} \left[\left(-1+\frac{z}{\sqrt{z^2+L^2} } \right)\hat{x}+ \left(\frac{L}{\sqrt{z^2+L^2} } \right)\hat{z} \right]$ .

For z ≫ L you expect it to look like a point charge $q=\lambda L:E\rightarrow \frac{1}{4\pi \varepsilon _0} \frac{\lambda L}{z^2}\hat{z}$ . It checks, for with z ≫L the $\hat{x}$ term → 0, and the $\hat{z}$ term $\rightarrow \frac{1}{4\pi \varepsilon _0}\frac{\lambda }{z}\frac{L}{z}\hat{z}$ .

Question 2.3: Find the electric field a distance z above one end of a stra...

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