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PhyHardAdv 2019 • Paper 2

Electric Dipole Fixed at Origin in Uniform Electric Field

An electric dipole having dipole moment p02(i^+j^)\frac{p_0}{\sqrt{2}}(\hat{i} + \hat{j}) is positioned at the origin OO fixed in space. It is placed within a uniform electric field of magnitude E0E_0. Provided that the electric potential remains constant along a circular path of radius RR centered at the origin as depicted in the figure, identify the valid statement(s): (Assume ϵ0\epsilon_0 is the permittivity of free space, and R≫R \gg dipole size)

Question Diagram 1

Options

A

R=(p04πϵ0E0)1/3R = \left(\frac{p_0}{4\pi \epsilon_0 E_0}\right)^{1/3}

Correct
B

Total electric field at point A is E⃗A=2E0(i^+j^)\vec{E}_A = \sqrt{2}E_0(\hat{i} + \hat{j})

C

Total electric field at point B is E⃗B=0\vec{E}_B = 0

Correct
D

The magnitude of the net electric field at any two locations on the circle is identical.

Topics & Concepts

Electric Charges and FieldsElectric PotentialElectric Field

Step-by-Step Solution

To solve this problem, we analyze the electric potential and electric field created by both the fixed electric dipole and the uniform external electric field.

1. Expression for Electric Potential

The dipole moment is given by: p⃗=p02(i^+j^)=p0n^p\vec{p} = \frac{p_0}{\sqrt{2}}(\hat{i} + \hat{j}) = p_0 \hat{n}_p where n^p=i^+j^2\hat{n}_p = \frac{\hat{i} + \hat{j}}{\sqrt{2}} is a unit vector along the 45∘45^\circ line relative to the positive xx-axis.

The electric potential at a point with position vector r⃗=Rr^\vec{r} = R \hat{r} on the circle of radius RR due to the dipole is: Vdipole(r⃗)=14πϵ0p⃗⋅r^R2=p04πϵ0R2(n^p⋅r^)V_{\text{dipole}}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{R^2} = \frac{p_0}{4\pi\epsilon_0 R^2} (\hat{n}_p \cdot \hat{r})

Let the uniform electric field be E⃗0\vec{E}_0. The potential associated with this uniform electric field is: Vext(r⃗)=−E⃗0⋅r⃗=−R(E⃗0⋅r^)V_{\text{ext}}(\vec{r}) = -\vec{E}_0 \cdot \vec{r} = -R (\vec{E}_0 \cdot \hat{r})

The total electric potential on the circle of radius RR is: V(r⃗)=Vdipole(r⃗)+Vext(r⃗)=(p04πϵ0R2n^p−RE⃗0)⋅r^V(\vec{r}) = V_{\text{dipole}}(\vec{r}) + V_{\text{ext}}(\vec{r}) = \left( \frac{p_0}{4\pi\epsilon_0 R^2} \hat{n}_p - R \vec{E}_0 \right) \cdot \hat{r}

Given that the potential remains constant along the circular path of radius RR, the potential must be independent of r^\hat{r}. This requires the vector coefficient of r^\hat{r} to vanish: p04πϵ0R2n^p−RE⃗0=0⃗\frac{p_0}{4\pi\epsilon_0 R^2} \hat{n}_p - R \vec{E}_0 = \vec{0}

From this, we find: E⃗0=p04πϵ0R3n^p\vec{E}_0 = \frac{p_0}{4\pi\epsilon_0 R^3} \hat{n}_p

Taking the magnitude on both sides (since ∣n^p∣=1|\hat{n}_p| = 1 and ∣E⃗0∣=E0|\vec{E}_0| = E_0): E0=p04πϵ0R3  ⟹  R=(p04πϵ0E0)1/3E_0 = \frac{p_0}{4\pi\epsilon_0 R^3} \implies R = \left(\frac{p_0}{4\pi\epsilon_0 E_0}\right)^{1/3}

Thus, Statement (A) is correct.

2. Total Electric Field on the Circle

The uniform electric field is: E⃗0=E0n^p=E02(i^+j^)\vec{E}_0 = E_0 \hat{n}_p = \frac{E_0}{\sqrt{2}}(\hat{i} + \hat{j})

The electric field of the dipole at any point r⃗=Rr^\vec{r} = R \hat{r} on the circle is given by: E⃗dipole(r⃗)=14πϵ0R3[3(p⃗⋅r^)r^−p⃗]=E0[3(n^p⋅r^)r^−n^p]\vec{E}_{\text{dipole}}(\vec{r}) = \frac{1}{4\pi\epsilon_0 R^3} \left[ 3(\vec{p} \cdot \hat{r})\hat{r} - \vec{p} \right] = E_0 \left[ 3(\hat{n}_p \cdot \hat{r})\hat{r} - \hat{n}_p \right]

The total electric field at position r⃗\vec{r} is: E⃗total(r⃗)=E⃗dipole(r⃗)+E⃗0=E0[3(n^p⋅r^)r^−n^p]+E0n^p=3E0(n^p⋅r^)r^\vec{E}_{\text{total}}(\vec{r}) = \vec{E}_{\text{dipole}}(\vec{r}) + \vec{E}_0 = E_0 \left[ 3(\hat{n}_p \cdot \hat{r})\hat{r} - \hat{n}_p \right] + E_0 \hat{n}_p = 3 E_0 (\hat{n}_p \cdot \hat{r})\hat{r}

3. Analysis of Options B, C, and D

  • At Point A: Point A is located along the direction of p⃗\vec{p} at θ=45∘\theta = 45^\circ, so r^A=n^p=i^+j^2\hat{r}_A = \hat{n}_p = \frac{\hat{i} + \hat{j}}{\sqrt{2}}. E⃗A=3E0(n^p⋅n^p)n^p=3E0n^p=3E02(i^+j^)\vec{E}_A = 3 E_0 (\hat{n}_p \cdot \hat{n}_p) \hat{n}_p = 3 E_0 \hat{n}_p = \frac{3 E_0}{\sqrt{2}}(\hat{i} + \hat{j}) Thus, Statement (B) is incorrect.

  • At Point B: Point B is located at θ=135∘\theta = 135^\circ, so r^B=−i^+j^2\hat{r}_B = \frac{-\hat{i} + \hat{j}}{\sqrt{2}}. Since r^B⊥n^p\hat{r}_B \perp \hat{n}_p, we have n^p⋅r^B=0\hat{n}_p \cdot \hat{r}_B = 0: E⃗B=3E0(0)r^B=0⃗\vec{E}_B = 3 E_0 (0) \hat{r}_B = \vec{0} Thus, Statement (C) is correct.

  • Magnitude of Net Electric Field: The magnitude of the total electric field at any point on the circle is: ∣E⃗total(r⃗)∣=3E0∣n^p⋅r^∣=3E0∣cos⁡(θ−45∘)∣|\vec{E}_{\text{total}}(\vec{r})| = 3 E_0 |\hat{n}_p \cdot \hat{r}| = 3 E_0 |\cos(\theta - 45^\circ)| Since this magnitude depends explicitly on the angle θ\theta, it varies along the circle (e.g., 3E03 E_0 at point A and 00 at point B). Thus, Statement (D) is incorrect.

Conclusion

The correct options are A and C.

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