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Question: An electric dipole is placed in a uniform electric field E which is normal to P. Find the direction ...

An electric dipole is placed in a uniform electric field E which is normal to P. Find the direction from the axis of dipole in which at some point magnitude of electric field can be zero

Answer

tan θ = √2

Explanation

Solution

Let the electric dipole be placed at the origin with its dipole moment vector p⃗\vec{p} along the x-axis, i.e., p⃗=pi^\vec{p} = p \hat{i}. The uniform electric field E⃗\vec{E} is normal to p⃗\vec{p}. Let's assume E⃗\vec{E} is along the y-axis. So, E⃗uniform=E0j^\vec{E}_{uniform} = E_0 \hat{j}, where E0E_0 is the magnitude of the uniform field.

The electric field due to the dipole at a point with polar coordinates (r,θ)(r, \theta), where θ\theta is the angle from the dipole axis (x-axis), is given by: E⃗dipole=kpr3[(3cos⁡2θ−1)i^+(3sin⁡θcos⁡θ)j^]\vec{E}_{dipole} = \frac{kp}{r^3} [(3 \cos^2 \theta - 1) \hat{i} + (3 \sin \theta \cos \theta) \hat{j}] where k=14πϵ0k = \frac{1}{4\pi\epsilon_0}.

The total electric field at the point is the vector sum of the dipole field and the uniform field: E⃗total=E⃗dipole+E⃗uniform\vec{E}_{total} = \vec{E}_{dipole} + \vec{E}_{uniform} E⃗total=kpr3[(3cos⁡2θ−1)i^+(3sin⁡θcos⁡θ)j^]+E0j^\vec{E}_{total} = \frac{kp}{r^3} [(3 \cos^2 \theta - 1) \hat{i} + (3 \sin \theta \cos \theta) \hat{j}] + E_0 \hat{j} E⃗total=kpr3(3cos⁡2θ−1)i^+(kpr3(3sin⁡θcos⁡θ)+E0)j^\vec{E}_{total} = \frac{kp}{r^3} (3 \cos^2 \theta - 1) \hat{i} + \left(\frac{kp}{r^3} (3 \sin \theta \cos \theta) + E_0\right) \hat{j}

For the total electric field to be zero at some point (r,θ)(r, \theta), both the i^\hat{i} and j^\hat{j} components must be zero. Component along i^\hat{i}: kpr3(3cos⁡2θ−1)=0\frac{kp}{r^3} (3 \cos^2 \theta - 1) = 0 Since k,p,rk, p, r are non-zero (we are looking for points at a finite distance where the dipole field exists), we must have: 3cos⁡2θ−1=03 \cos^2 \theta - 1 = 0 3cos⁡2θ=13 \cos^2 \theta = 1 cos⁡2θ=13\cos^2 \theta = \frac{1}{3} cos⁡θ=±13\cos \theta = \pm \frac{1}{\sqrt{3}}

Component along j^\hat{j}: kpr3(3sin⁡θcos⁡θ)+E0=0\frac{kp}{r^3} (3 \sin \theta \cos \theta) + E_0 = 0 kpr3(3sin⁡θcos⁡θ)=−E0\frac{kp}{r^3} (3 \sin \theta \cos \theta) = -E_0

Now we substitute the possible values of cos⁡θ\cos \theta into the second equation.

Case 1: cos⁡θ=13\cos \theta = \frac{1}{\sqrt{3}} The second equation becomes: kpr3(3sin⁡θ13)=−E0\frac{kp}{r^3} (3 \sin \theta \frac{1}{\sqrt{3}}) = -E_0 kpr33sin⁡θ=−E0\frac{kp}{r^3} \sqrt{3} \sin \theta = -E_0 Since k,p,r,3,E0k, p, r, \sqrt{3}, E_0 are positive quantities, sin⁡θ\sin \theta must be negative. If cos⁡θ=13\cos \theta = \frac{1}{\sqrt{3}}, then sin⁡2θ=1−cos⁡2θ=1−13=23\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{3} = \frac{2}{3}. Since sin⁡θ\sin \theta must be negative, sin⁡θ=−23\sin \theta = -\sqrt{\frac{2}{3}}. This corresponds to an angle θ\theta in the 4th quadrant (where cos⁡θ>0\cos \theta > 0 and sin⁡θ<0\sin \theta < 0). The direction is given by tan⁡θ=sin⁡θcos⁡θ=−2/31/3=−2\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\sqrt{2/3}}{1/\sqrt{3}} = -\sqrt{2}. For this direction, the distance rr is given by kpr33(−23)=−E0\frac{kp}{r^3} \sqrt{3} (-\sqrt{\frac{2}{3}}) = -E_0, which simplifies to kpr32=E0\frac{kp}{r^3} \sqrt{2} = E_0. r3=2kpE0r^3 = \frac{\sqrt{2} kp}{E_0}. Since k,p,E0k, p, E_0 are positive, rr is real and positive, so points exist in this direction.

Case 2: cos⁡θ=−13\cos \theta = -\frac{1}{\sqrt{3}} The second equation becomes: kpr3(3sin⁡θ(−13))=−E0\frac{kp}{r^3} (3 \sin \theta (-\frac{1}{\sqrt{3}})) = -E_0 −kpr33sin⁡θ=−E0-\frac{kp}{r^3} \sqrt{3} \sin \theta = -E_0 kpr33sin⁡θ=E0\frac{kp}{r^3} \sqrt{3} \sin \theta = E_0 Since k,p,r,3,E0k, p, r, \sqrt{3}, E_0 are positive, sin⁡θ\sin \theta must be positive. If cos⁡θ=−13\cos \theta = -\frac{1}{\sqrt{3}}, then sin⁡2θ=1−cos⁡2θ=1−13=23\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{3} = \frac{2}{3}. Since sin⁡θ\sin \theta must be positive, sin⁡θ=23\sin \theta = \sqrt{\frac{2}{3}}. This corresponds to an angle θ\theta in the 2nd quadrant (where cos⁡θ<0\cos \theta < 0 and sin⁡θ>0\sin \theta > 0). The direction is given by tan⁡θ=sin⁡θcos⁡θ=2/3−1/3=−2\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{2/3}}{-1/\sqrt{3}} = -\sqrt{2}. For this direction, the distance rr is given by kpr33(23)=E0\frac{kp}{r^3} \sqrt{3} (\sqrt{\frac{2}{3}}) = E_0, which simplifies to kpr32=E0\frac{kp}{r^3} \sqrt{2} = E_0. r3=2kpE0r^3 = \frac{\sqrt{2} kp}{E_0}. This is the same distance as in Case 1, and it is real and positive.

Both cases give the same condition for the direction: tan⁡θ=−2\tan \theta = -\sqrt{2}. The angle θ\theta is the angle from the dipole axis. The directions are θ=arctan⁡(−2)\theta = \arctan(-\sqrt{2}) in the 2nd and 4th quadrants.

If the uniform field was E⃗uniform=−E0j^\vec{E}_{uniform} = -E_0 \hat{j}, then the condition for zero field would be E⃗dipole=−E⃗uniform=E0j^\vec{E}_{dipole} = - \vec{E}_{uniform} = E_0 \hat{j}. The i^\hat{i} component condition remains 3cos⁡2θ−1=03 \cos^2 \theta - 1 = 0, so cos⁡θ=±13\cos \theta = \pm \frac{1}{\sqrt{3}}. The j^\hat{j} component condition becomes kpr3(3sin⁡θcos⁡θ)=E0\frac{kp}{r^3} (3 \sin \theta \cos \theta) = E_0. If cos⁡θ=13\cos \theta = \frac{1}{\sqrt{3}}, then kpr33sin⁡θ=E0\frac{kp}{r^3} \sqrt{3} \sin \theta = E_0. This requires sin⁡θ>0\sin \theta > 0, so sin⁡θ=23\sin \theta = \sqrt{\frac{2}{3}}. tan⁡θ=2\tan \theta = \sqrt{2} (1st quadrant). If cos⁡θ=−13\cos \theta = -\frac{1}{\sqrt{3}}, then −kpr33sin⁡θ=E0-\frac{kp}{r^3} \sqrt{3} \sin \theta = E_0. This requires sin⁡θ<0\sin \theta < 0, so sin⁡θ=−23\sin \theta = -\sqrt{\frac{2}{3}}. tan⁡θ=2\tan \theta = \sqrt{2} (3rd quadrant). In this case, the directions are given by tan⁡θ=2\tan \theta = \sqrt{2}.

The question states the uniform field is normal to P. It does not specify the direction (e.g., +y or -y). The problem is asking for the direction, implying there is a unique answer or a set of directions irrespective of the sign of E0E_0. However, the direction of the zero field point depends on the direction of the external field. Let's re-read carefully: "An electric dipole is placed in a uniform electric field E which is normal to P." Let p⃗\vec{p} be along the x-axis. E⃗\vec{E} is along the y-axis (either +E0j^+E_0 \hat{j} or −E0j^-E_0 \hat{j}). If E⃗=E0j^\vec{E} = E_0 \hat{j}, zero field points exist at angles θ\theta from the dipole axis such that tan⁡θ=−2\tan \theta = -\sqrt{2}. If E⃗=−E0j^\vec{E} = -E_0 \hat{j}, zero field points exist at angles θ\theta from the dipole axis such that tan⁡θ=2\tan \theta = \sqrt{2}.

The question asks "Find the direction from the axis of dipole in which at some point magnitude of electric field can be zero". It asks for the direction. This suggests the answer should be independent of the sign of E0E_0. Let's consider the angle relative to the dipole axis. The magnitude of the angle is what defines the direction from the axis. In the first case (E⃗=E0j^\vec{E} = E_0 \hat{j}), the angles are θ1=arctan⁡(−2)\theta_1 = \arctan(-\sqrt{2}) in the 2nd quadrant and θ2=arctan⁡(−2)\theta_2 = \arctan(-\sqrt{2}) in the 4th quadrant. The angle relative to the axis (x-axis) is ∣θ1∣|\theta_1| or ∣θ2∣|\theta_2|. Let α\alpha be the angle such that tan⁡α=2\tan \alpha = \sqrt{2} and 0<α<π/20 < \alpha < \pi/2. Then the angles are π−α\pi - \alpha and 2π−α2\pi - \alpha (or −α-\alpha). The angle from the axis is α\alpha. In the second case (E⃗=−E0j^\vec{E} = -E_0 \hat{j}), the angles are θ3=arctan⁡(2)\theta_3 = \arctan(\sqrt{2}) in the 1st quadrant and θ4=arctan⁡(2)\theta_4 = \arctan(\sqrt{2}) in the 3rd quadrant. These angles are α\alpha and π+α\pi + \alpha. The angle from the axis is α\alpha. In both scenarios, the magnitude of the angle from the dipole axis has a tangent of 2\sqrt{2}. So, the direction relative to the dipole axis where the field can be zero is given by an angle θ\theta such that ∣tan⁡θ∣=2|\tan \theta| = \sqrt{2}.

The question phrasing "Find the direction from the axis of dipole" implies the angle relative to the axis. Let α\alpha be the angle from the dipole axis. Then tan⁡α=2\tan \alpha = \sqrt{2}. The angle α\alpha is arctan⁡(2)\arctan(\sqrt{2}).

Final check: If cos⁡2θ=1/3\cos^2 \theta = 1/3, then sin⁡2θ=2/3\sin^2 \theta = 2/3. The radial component of dipole field is Er=2kpcos⁡θr3E_r = \frac{2kp \cos \theta}{r^3}. The tangential component of dipole field is Eθ=kpsin⁡θr3E_\theta = \frac{kp \sin \theta}{r^3}. The direction of E⃗dipole\vec{E}_{dipole} is at an angle ϕ\phi from the radial direction such that tan⁡ϕ=EθEr=kpsin⁡θ/r32kpcos⁡θ/r3=12tan⁡θ\tan \phi = \frac{E_\theta}{E_r} = \frac{kp \sin \theta / r^3}{2kp \cos \theta / r^3} = \frac{1}{2} \tan \theta. We require E⃗dipole\vec{E}_{dipole} to be opposite to E⃗uniform\vec{E}_{uniform}. If E⃗uniform\vec{E}_{uniform} is along +y, E⃗dipole\vec{E}_{dipole} is along -y. If E⃗uniform\vec{E}_{uniform} is along -y, E⃗dipole\vec{E}_{dipole} is along +y. The y-axis is perpendicular to the dipole axis (x-axis). The line where cos⁡2θ=1/3\cos^2 \theta = 1/3 is where the axial component of the dipole field is zero. On this line, the dipole field is purely tangential. The direction of the tangential component is perpendicular to the radial vector r⃗\vec{r}. If θ\theta is in the 2nd quadrant (cos⁡θ<0,sin⁡θ>0\cos \theta < 0, \sin \theta > 0), cos⁡θ=−1/3,sin⁡θ=2/3\cos \theta = -1/\sqrt{3}, \sin \theta = \sqrt{2/3}. Eθ=kpr32/3E_\theta = \frac{kp}{r^3} \sqrt{2/3}. EθE_\theta is positive, so it's in the direction of increasing θ\theta. This direction is roughly towards -x, +y. The angle θ\theta is in Q2. The tangential direction θ^=−sin⁡θi^+cos⁡θj^=−2/3i^−1/3j^\hat{\theta} = -\sin \theta \hat{i} + \cos \theta \hat{j} = -\sqrt{2/3} \hat{i} - 1/\sqrt{3} \hat{j}. This is in Q3. Something is wrong here.

Let's use the Cartesian components derived earlier. E⃗dipole=kpr3[(3cos⁡2θ−1)i^+(3sin⁡θcos⁡θ)j^]\vec{E}_{dipole} = \frac{kp}{r^3} [(3 \cos^2 \theta - 1) \hat{i} + (3 \sin \theta \cos \theta) \hat{j}]. The condition 3cos⁡2θ−1=03 \cos^2 \theta - 1 = 0 means the x-component of E⃗dipole\vec{E}_{dipole} is zero. So, on the line cos⁡2θ=1/3\cos^2 \theta = 1/3, E⃗dipole\vec{E}_{dipole} is purely along the y-direction (positive or negative). E⃗dipole=kpr3(3sin⁡θcos⁡θ)j^\vec{E}_{dipole} = \frac{kp}{r^3} (3 \sin \theta \cos \theta) \hat{j}. For the total field to be zero, this must be equal to −E⃗uniform-\vec{E}_{uniform}. If E⃗uniform=E0j^\vec{E}_{uniform} = E_0 \hat{j}, we need kpr3(3sin⁡θcos⁡θ)=−E0\frac{kp}{r^3} (3 \sin \theta \cos \theta) = -E_0. If cos⁡θ=1/3\cos \theta = 1/\sqrt{3}, sin⁡θ=±2/3\sin \theta = \pm \sqrt{2/3}. We need sin⁡θ\sin \theta to be negative, sin⁡θ=−2/3\sin \theta = -\sqrt{2/3}. This is the 4th quadrant. tan⁡θ=−2\tan \theta = -\sqrt{2}. If cos⁡θ=−1/3\cos \theta = -1/\sqrt{3}, sin⁡θ=±2/3\sin \theta = \pm \sqrt{2/3}. We need sin⁡θ\sin \theta to be positive, sin⁡θ=2/3\sin \theta = \sqrt{2/3}. This is the 2nd quadrant. tan⁡θ=−2\tan \theta = -\sqrt{2}.

If E⃗uniform=−E0j^\vec{E}_{uniform} = -E_0 \hat{j}, we need kpr3(3sin⁡θcos⁡θ)=E0\frac{kp}{r^3} (3 \sin \theta \cos \theta) = E_0. If cos⁡θ=1/3\cos \theta = 1/\sqrt{3}, sin⁡θ=±2/3\sin \theta = \pm \sqrt{2/3}. We need sin⁡θ\sin \theta to be positive, sin⁡θ=2/3\sin \theta = \sqrt{2/3}. This is the 1st quadrant. tan⁡θ=2\tan \theta = \sqrt{2}. If cos⁡θ=−1/3\cos \theta = -1/\sqrt{3}, sin⁡θ=±2/3\sin \theta = \pm \sqrt{2/3}. We need sin⁡θ\sin \theta to be negative, sin⁡θ=−2/3\sin \theta = -\sqrt{2/3}. This is the 3rd quadrant. tan⁡θ=2\tan \theta = \sqrt{2}.

The directions from the dipole axis where the field can be zero are those where tan⁡θ=2\tan \theta = \sqrt{2} or tan⁡θ=−2\tan \theta = -\sqrt{2}. This means the angle θ\theta satisfies tan⁡2θ=2\tan^2 \theta = 2. The question asks for "the direction from the axis". This usually refers to the angle relative to the axis. The magnitude of the angle whose tangent is 2\sqrt{2} is arctan⁡(2)\arctan(\sqrt{2}).

The angle θ\theta is measured from the dipole axis. The directions are given by tan⁡θ=±2\tan \theta = \pm \sqrt{2}. The angle relative to the axis is ∣θ∣|\theta|. If tan⁡θ=2\tan \theta = \sqrt{2}, the angles are arctan⁡(2)\arctan(\sqrt{2}) (1st quadrant) and π+arctan⁡(2)\pi + \arctan(\sqrt{2}) (3rd quadrant). The angle from the axis is arctan⁡(2)\arctan(\sqrt{2}). If tan⁡θ=−2\tan \theta = -\sqrt{2}, the angles are π−arctan⁡(2)\pi - \arctan(\sqrt{2}) (2nd quadrant) and 2π−arctan⁡(2)2\pi - \arctan(\sqrt{2}) (4th quadrant). The angle from the axis is π−(π−arctan⁡(2))=arctan⁡(2)\pi - (\pi - \arctan(\sqrt{2})) = \arctan(\sqrt{2}) or ∣−arctan⁡(2)∣=arctan⁡(2)|-\arctan(\sqrt{2})| = \arctan(\sqrt{2}).

In all cases, the angle from the dipole axis (x-axis) is α\alpha such that tan⁡α=2\tan \alpha = \sqrt{2}.

The final answer is tan⁡θ=2\tan \theta = \sqrt{2}, where θ\theta is the angle from the dipole axis.