Question

An electric dipole is placed along the \(\mathrm{x}\) -axis at the origin o. \(\mathrm{A}\) point \(P\) is at a distance of \(20 \mathrm{~cm}\) from this origin such that OP makes an angle \((\pi / 3)\) with the x-axis. If the electric field at P makes an angle \(\theta\) with the x-axis, the value of \(\theta\) would be \(\ldots \ldots \ldots\) (A) \((\pi / 3)+\tan ^{-1}(\sqrt{3} / 2)\) (B) \((\pi / 3)\) (C) \((2 \pi / 3)\) (D) \(\tan ^{-1}(\sqrt{3} / 2)\)

Short answer

The value of θ would be \(\tan^{-1}(\sqrt{3})\). The correct answer is (D).

Step-by-step solution

Calculate the electric field components

To find the electric field components, we first need to rewrite the electric dipole in its Cartesian coordinates. We know that P is 20 cm from the origin at an angle of π/3. So, the coordinates of P are: P(x, y) = \((20 \cos(\pi / 3), 20 \sin(\pi / 3))\) Given that the electric field formula for a dipole along the x and y axes is: \(E_x = \frac{2 p_x}{4\pi\epsilon_0} \frac{x}{(x^2 + y^2)^{3/2}}\) \(E_y = \frac{p_x}{4\pi\epsilon_0} \frac{y}{(x^2 + y^2)^{3/2}}\) We can find the electric field along the x and y axes at point P using the electric dipole moment p_x: \(E_x = \frac{2 p_x}{4\pi\epsilon_0} \frac{(20\cos(\pi / 3))}{((20\cos(\pi / 3))^2 + (20\sin(\pi / 3))^2)^{3/2}}\) \(E_y = \frac{p_x}{4\pi\epsilon_0} \frac{(20\sin(\pi / 3))}{((20\cos(\pi / 3))^2 + (20\sin(\pi / 3))^2)^{3/2}}\)

Find the overall electric field at point P

The overall electric field at point P can be found by adding the electric field components along the x-axis (E_x) and the y-axis (E_y). This gives us: \(E_P = \sqrt{E_x^2 + E_y^2}\) Substituting the values for E_x and E_y, we get: \(E_P = \sqrt{ \frac{(2\cos(\pi / 3))^2p_x^2}{(4\pi\epsilon_0)^2(20^2\cos^2(\pi / 3) + 20^2\sin^2(\pi / 3))^2} + \frac{(\sin(\pi / 3))^2p_x^2}{(4\pi\epsilon_0)^2(20^2\cos^2(\pi / 3) + 20^2\sin^2(\pi / 3))^2}}\)

Calculate the angle θ

Now, we can find the angle θ, that the electric field at P makes with the x-axis by using the tangent function: \(\tan(\theta) = \frac{E_y}{E_x}\) So, the angle θ is given by: \(\theta = \tan^{-1} \left( \frac{E_y}{E_x} \right)\) Substituting the values for E_x and E_y, we get: \(\theta = \tan^{-1} \left( \frac{\frac{p_x}{4\pi\epsilon_0} \frac{(20\sin(\pi / 3))}{((20\cos(\pi / 3))^2 + (20\sin(\pi / 3))^2)^{3/2}}}{\frac{2 p_x}{4\pi\epsilon_0} \frac{(20\cos(\pi / 3))}{((20\cos(\pi / 3))^2 + (20\sin(\pi / 3))^2)^{3/2}}} \right)\) Simplifying this expression, we get: \(\theta = \tan^{-1} \left( \frac{\sin(\pi / 3)}{2\cos(\pi / 3)} \right) = \tan^{-1} \left( \frac{\sqrt{3} / 2}{2(1 / 2)} \right) = \tan^{-1}(\sqrt{3})\) Therefore, the value of θ would be \(\tan^{-1}(\sqrt{3})\). The correct answer is (D).