Q58E Consider the electric dipole of ... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 4: Q58E (page 717)

Consider the electric dipole of Example 21.14. (a) Derive an expression for the magnitude of the electric field produced by the dipole at a point on the x-axis in Fig. 21.33. What is the direction of this electric field? (b) How does the electric field at points on the x-axis depend on x when x is very large?

Short Answer

Expert verified

(a) The magnitude of the electric field produced by the dipole at a point on the x axis is $E = \frac{鈭� k q d}{{[(x)^{2} + (d / 2)^{2}]}^{3 / 2}}$ . The direction of this electric field is in $- \hat{j}$ direction.

(b) Electric field at points on the x-axis depend on x when x is very large as $E = \frac{鈭� k q d}{(x)^{3}}$ .

Step by step solution

Step 1:

According to super position law the electric field component in x direction at point x is given by

$E_{x} = 鈭� E_{鈭�} \cos 鈦� 胃 + E_{+} \cos 鈦� 胃 = 0$

Because $E_{-} = E_{+}$ in magnitude

The y component is given by

$\begin{aligned} E_{y} = 鈭� E_{鈭�} \sin 鈦� 胃 + E_{+} \sin 鈦� 胃 \\ = 鈭� \frac{2 k q}{r^{2}} 脳 \frac{d}{2 r} \\ = \frac{鈭� k q d}{{[(x)^{2} + (d / 2)^{2}]}^{3 / 2}} \end{aligned}$

The total field E is given by

$\begin{aligned} E = \frac{鈭� k q d}{{[(x)^{2} + (d / 2)^{2}]}^{3 / 2}} \end{aligned}$

And the direction is $- \hat{j}$

Therefore, the magnitude of the electric field produced by the dipole at a point on the x axis is $E = \frac{鈭� k q d}{{[(x)^{2} + (d / 2)^{2}]}^{3 / 2}}$ and the direction of this electric field is in $- \hat{j}$ direction.

Step 2:

X is very large compared with d/2:

$(x)^{2} + (d / 2)^{2} 鈮� x^{2}$

Hence total field is

$\begin{aligned} E = \frac{鈭� k q d}{{[(x)^{2}]}^{3 / 2}} \\ = \frac{鈭� k q d}{(x)^{3}} \end{aligned}$

Thus electric field at points on the x-axis depend on x when x is very large as.

$E = \frac{鈭� k q d}{(x)^{3}}$

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