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

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

Step by step solution

01

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: \(E_x = \frac{2 p_x}{4\pi\epsilon_0} \frac{(20\cos(\pi / 3))}{((20\cos(\pi / 3))^2 + (20\sin(\pi / 3))^2)^{3/2}}\) \(E_y = \frac{p_x}{4\pi\epsilon_0} \frac{(20\sin(\pi / 3))}{((20\cos(\pi / 3))^2 + (20\sin(\pi / 3))^2)^{3/2}}\)
02

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: \(E_P = \sqrt{ \frac{(2\cos(\pi / 3))^2p_x^2}{(4\pi\epsilon_0)^2(20^2\cos^2(\pi / 3) + 20^2\sin^2(\pi / 3))^2} + \frac{(\sin(\pi / 3))^2p_x^2}{(4\pi\epsilon_0)^2(20^2\cos^2(\pi / 3) + 20^2\sin^2(\pi / 3))^2}}\)
03

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).

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