Question 2.1: Find the electric field a distance z above the midpoint betw...

Introduction to Electrodynamics 4th. Edition [EXP-2861]

Find the electric field a distance z above the midpoint between two equal charges (q), a distance d apart (Fig. 2.4a).

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

Two equal charges (q)
Distance (d) between the charges
Distance (z) above the midpoint

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Step 1:

In order to find the electric field at a distance z above the midpoint between two equal charges, we need to first consider the electric fields of each charge separately. Let's call the electric field of the left charge alone as E1 and that of the right charge alone as E2.

Step 2:

When we add the electric fields E1 and E2 together (vectorially), the horizontal components cancel out and only the vertical components contribute to the resulting electric field.

Step 3:

The resulting electric field, denoted as Ez, can be calculated using the formula: Ez = 2/(4πε0) (1/η^2) q * cosθ, where η is a quantity defined as η = √(z^2 + (d/2)^2) and cosθ = z/η.

Step 4:

In the formula, q represents the magnitude of the charges, and ε0 is the permittivity of free space.

Step 5:

To simplify the formula, we can rewrite η^2 as (z^2 + (d/2)^2) and simplify the expression.

Step 6:

Finally, we arrive at the formula for Ez: Ez = (1/4πε0) * (2qz) / (z^2 + (d/2)^2)^(3/2)z.

Step 7:

It is important to note that when the distance z is much larger than the distance d, the electric field Ez approaches the electric field of a single charge 2q. In this case, we can simplify the formula for Ez to Ez = (1/4πε0) * (2q) / z^2 z.

Step 8:

This relationship can be verified by setting d to zero in the formula, which would mean that the charges are very close together.

Step 9:

It is important to understand the step-by-step process of solving the problem in order to accurately determine the electric field at a distance z above the midpoint of two equal charges separated by a distance d.

Final Answer

Let $E_{1}$ be the field of the left charge alone, and $E_{2}$ that of the right charge alone (Fig. 2.4b). Adding them (vectorially), the horizontal components cancel and the vertical components conspire:

E_{z}=2\frac{1}{4\pi \varepsilon _{0} }\frac{q}{η ^{2}} \cos \theta

Here $η =\sqrt{z^{2}+(d/2)^{2}}$ and $\cos \theta =z/η$ ,so

E=\frac{1}{4\pi \varepsilon _{0}}\frac{2qz}{[z^{2}+(d/2)^{2}]^{3/2}}\hat{z}

Check: When $z\gg d$ you’re so far away that it just looks like a single charge 2q, so the field should reduce to $E=\frac{1}{4\pi \varepsilon _{0}}\frac{2q}{z^{2}}\hat{z}$ .And it does (just set d → 0 in the formula).

Question 2.1: Find the electric field a distance z above the midpoint betw...

Final Answer

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