A dipole consists of two equal and opposite point charges \((\pm q)\) separated by a distance \(d\) (a) Write an expression for the magnitude of the electric field at a point \((x, 0)\) a large distance \((x \gg d)\) from the midpoint of the charges on a line perpendicular to the dipole axis. [Hint: Use small angle approximations.] (b) Give the direction of the field for \(x>0\) and for \(x<0\).

The electric field from a point charge is given by the expression: \(E = \frac{k|q|}{r^2}\), where \(k\) is the Coulomb's constant (\(8.99\times10^{9} Nm^2/C^2\)), \(q\) is the charge, and \(r\) is the distance from the point charge to the point of interest.

Let's consider the dipole with positive charge \(q\) located at \((0, d/2)\) and negative charge \(-q\) located at \((0, -d/2)\). The distance from each charge to the point \((x, 0)\) is given by: \(r_+ = \sqrt{x^2 + (d/2)^2}\) for the positive charge \(r_- = \sqrt{x^2 + (d/2)^2}\) for the negative charge Since the distances are the same, the magnitudes of the electric fields from each charge at point \((x, 0)\) are also equal.

The electric field vectors from each charge to the point \((x, 0)\) can be written as: \(\vec{E}_+ = E_+ \cos\theta \hat{i} - E_+ \sin\theta \hat{j}\) \(\vec{E}_- = -E_- \cos\theta \hat{i} - E_- \sin\theta \hat{j}\) Where \(E_+ = E_- = \frac{k|q|}{r^2}\), and \(\theta\) is the angle between the electric field vector and the horizontal axis.

Adding the electric field vectors, we obtain the net electric field: \(\vec{E} = (\frac{2kq\cos\theta}{r^2})\hat{i} - (\frac{2kq\sin\theta}{r^2})\hat{j}\)

Since \(x \gg d\), the angle \(\theta\) is very small. The small angle approximations are: \(\cos\theta \approx 1\) and \(\sin\theta \approx \frac{d/2}{r}\). Substituting these approximations, we get: \(\vec{E} = (\frac{2kq}{r^2})\hat{i} - (\frac{kd}{r^3})\hat{j}\). But \(r^2 \approx x^2\), so the magnitude of the electric field at \((x, 0)\) is approximately: \(E = \sqrt{(\frac{2kq}{x^2})^2 + (\frac{kd}{x^3})^2}\) (a) Since \(x^2 \gg d^2\), we can ignore the second term in the magnitude under the square root sign, and obtain: \(E \approx \frac{2kq}{x^2}\) (b) The direction of the electric field depends on the sign of \(x\). For \(x > 0\), the electric field vector \(\vec{E}\) points to the right and downward (positive \(\hat{i}\) direction and negative \(\hat{j}\) direction). For \(x < 0\), the electric field vector \(\vec{E}\) points to the left and downward (negative \(\hat{i}\) direction and negative \(\hat{j}\) direction).