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Question: Consider two elongated uniform charge carriers of infinite extent having identical magnitude line ch...

Consider two elongated uniform charge carriers of infinite extent having identical magnitude line charge distributions but with opposing polarity orientations as shown. The electric field at any coordinate position along the horizontal reference axis (x-axis) for positive values of x-coordinate follows the directional unit vector

A

cos\theta \hat{i} + sin\theta \hat{j}

B

\hat{i}

C

\hat{j}

D

-sin\theta \hat{i} + sin\theta \hat{j}

Answer

\hat{j}

Explanation

Solution

The electric field at a point on the x-axis (x>0x>0) due to the positive line charge is directed perpendicular to the line, away from it. The distance from (x,0)(x,0) to this line is r=xsin⁡θr = x \sin \theta. The magnitude of the field is E1=λ2πϵ0r=λ2πϵ0xsin⁡θE_1 = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{\lambda}{2\pi\epsilon_0 x \sin \theta}. The direction of this field vector makes an angle of 90∘+θ90^\circ + \theta with the positive x-axis. Thus, E⃗1=E1(cos⁡(90∘+θ)i^+sin⁡(90∘+θ)j^)=E1(−sin⁡θi^+cos⁡θj^)\vec{E}_1 = E_1 (\cos(90^\circ+\theta) \hat{i} + \sin(90^\circ+\theta) \hat{j}) = E_1 (-\sin \theta \hat{i} + \cos \theta \hat{j}).

The electric field at the same point (x,0)(x,0) due to the negative line charge is directed perpendicular to the line, towards it. The distance to this line is also r=xsin⁡θr = x \sin \theta. The magnitude of the field is E2=λ2πϵ0r=λ2πϵ0xsin⁡θE_2 = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{\lambda}{2\pi\epsilon_0 x \sin \theta}. The line y=−(tan⁡θ)xy = -(\tan \theta) x makes an angle of −θ-\theta with the x-axis. The field points towards this line. The total electric field is E⃗=E⃗1+E⃗2\vec{E} = \vec{E}_1 + \vec{E}_2. E⃗=λ2πϵ0xsin⁡θ(−sin⁡θi^+cos⁡θj^)+λ2πϵ0xsin⁡θ(sin⁡θi^+cos⁡θj^)\vec{E} = \frac{\lambda}{2\pi\epsilon_0 x \sin \theta} (-\sin \theta \hat{i} + \cos \theta \hat{j}) + \frac{\lambda}{2\pi\epsilon_0 x \sin \theta} (\sin \theta \hat{i} + \cos \theta \hat{j}) E⃗=λ2πϵ0xsin⁡θ(−sin⁡θ+sin⁡θ)i^+λ2πϵ0xsin⁡θ(cos⁡θ+cos⁡θ)j^\vec{E} = \frac{\lambda}{2\pi\epsilon_0 x \sin \theta} (-\sin \theta + \sin \theta) \hat{i} + \frac{\lambda}{2\pi\epsilon_0 x \sin \theta} (\cos \theta + \cos \theta) \hat{j} E⃗=λ2πϵ0xsin⁡θ(2cos⁡θj^)=λcot⁡θπϵ0xj^\vec{E} = \frac{\lambda}{2\pi\epsilon_0 x \sin \theta} (2 \cos \theta \hat{j}) = \frac{\lambda \cot \theta}{\pi\epsilon_0 x} \hat{j} The electric field is in the +j^+\hat{j} direction. The question asks for the directional unit vector.