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 $E = \frac{- k q d}{{[(x)^{2} + (d / 2)^{2}]}^{3 / 2}}$ . 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 $E = \frac{- k q d}{(x)^{3}}$ .

Step by step solution

Step 1:

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

$E_{x} = - E_{-} \cos θ + E_{+} \cos θ = 0$

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

The y component is given by

$\begin{aligned} E_{y} = - E_{-} \sin θ + E_{+} \sin θ \\ = - \frac{2 k q}{r^{2}} \times \frac{d}{2 r} \\ = \frac{- k q d}{{[(x)^{2} + (d / 2)^{2}]}^{3 / 2}} \end{aligned}$

The total field E is given by

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

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 $E = \frac{- k q d}{{[(x)^{2} + (d / 2)^{2}]}^{3 / 2}}$ and the direction of this electric field is in $- \hat{j}$ direction.