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Electrostatics question

2019 · Shift 2 · Q37
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Electrostatics question

2019 · Shift 2 · Q37

JEE AdvancedPhysicsElectrostaticsMultiple correct+4 / −1
An electric dipole with dipole moment p02(i^+j^){{{p_0}} \over {\sqrt 2 }}(\widehat i + \widehat j)2​p0​​(i+j​) is held fixed at the origin O in the presence of a uniform electric field of magnitude E0. JEE Advanced 2019 Paper 2 Offline Physics - Electrostatics Question 37 English If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are, (∈\in∈ 0 is the permittivity of the free space, R >> dipole size)
  1. A
    The magnitude of total electric field on any two points of the circle will be same.
  2. B
    Total electric field at point B is E→B{\overrightarrow E _B}EB​ = 0
  3. C
    R=(p04π∈0E0)1/3R = {\left( {{{{p_0}} \over {4\pi { \in _0}{E_0}}}} \right)^{1/3}}R=(4π∈0​E0​p0​​)1/3
  4. D
    Total electric field at point A is E→A=2E0(i^+j^){\overrightarrow E _A} = \sqrt 2 {E_0}(\widehat i + \widehat j)EA​=2​E0​(i+j​)
View written solutionFree

Correct answer: B, C

  1. Given data

    The dipole moment is

    p⃗=p02(i^+j^)\vec p = \frac{p_0}{\sqrt 2}(\hat i+\hat j)p​=2​p0​​(i^+j^​)

    so it points along the direction i^+j^\hat i+\hat ji^+j^​ (i.e. along 45∘45^\circ45∘).

    A uniform electric field of magnitude E0E_0E0​ is also present. We are told that the potential is constant on a circle of radius RRR centered at the origin.

  1. Potential due to dipole + uniform field

    For a point at position vector r⃗\vec rr with r=Rr=Rr=R, the dipole potential is

    Vd=14πε0p⃗⋅r⃗R3V_d = \frac{1}{4\pi\varepsilon_0}\frac{\vec p\cdot \vec r}{R^3}Vd​=4πε0​1​R3p​⋅r​

    The potential due to a uniform field E⃗0\vec E_0E0​ is

    Vu=−E⃗0⋅r⃗V_u = -\vec E_0\cdot \vec rVu​=−E0​⋅r

    Hence total potential on the circle is

    V=14πε0p⃗⋅r⃗R3−E⃗0⋅r⃗V = \frac{1}{4\pi\varepsilon_0}\frac{\vec p\cdot \vec r}{R^3} - \vec E_0\cdot \vec rV=4πε0​1​R3p​⋅r​−E0​⋅r

    Since this is constant for all points on the circle, the coefficient of r⃗\vec rr must vanish:

    14πε0R3p⃗−E⃗0=0\frac{1}{4\pi\varepsilon_0R^3}\vec p - \vec E_0 = 04πε0​R31​p​−E0​=0

    therefore

    E⃗0=14πε0R3p⃗\vec E_0 = \frac{1}{4\pi\varepsilon_0R^3}\vec pE0​=4πε0​R31​p​

    So the uniform field must be parallel to p⃗\vec pp​, with magnitude

    E0=p04πε0R3E_0 = \frac{p_0}{4\pi\varepsilon_0R^3}E0​=4πε0​R3p0​​

    because ∣p⃗∣=p0|\vec p|=p_0∣p​∣=p0​.

    Hence

    R=(p04πε0E0)1/3R=\left(\frac{p_0}{4\pi\varepsilon_0E_0}\right)^{1/3}R=(4πε0​E0​p0​​)1/3

    so Option C is correct.

  1. Direction of the uniform electric field

    Since E⃗0∥p⃗\vec E_0 \parallel \vec pE0​∥p​,

    E⃗0=E02(i^+j^)\vec E_0 = \frac{E_0}{\sqrt2}(\hat i+\hat j)E0​=2​E0​​(i^+j^​)

  1. Electric field due to dipole on the circle

    For a dipole, the field at position r^\hat rr^ is

    E⃗d=14πε0R3[3(p⃗⋅r^)r^−p⃗]\vec E_d = \frac{1}{4\pi\varepsilon_0R^3}\left[3(\vec p\cdot \hat r)\hat r-\vec p\right]Ed​=4πε0​R31​[3(p​⋅r^)r^−p​]

    Using

    14πε0R3p⃗=E⃗0\frac{1}{4\pi\varepsilon_0R^3}\vec p = \vec E_04πε0​R31​p​=E0​

    we get

    E⃗d=3(E⃗0⋅r^)r^−E⃗0\vec E_d = 3(\vec E_0\cdot \hat r)\hat r-\vec E_0Ed​=3(E0​⋅r^)r^−E0​

    Therefore total field is

    E⃗=E⃗d+E⃗0=3(E⃗0⋅r^)r^\vec E = \vec E_d+\vec E_0 = 3(\vec E_0\cdot \hat r)\hat rE=Ed​+E0​=3(E0​⋅r^)r^

    So the total field depends on the point on the circle through r^\hat rr^.

  1. Check option B

    From the figure (standard labeling for this question), point BBB lies on the diameter perpendicular to p⃗\vec pp​ (and hence perpendicular to E⃗0\vec E_0E0​). Thus at BBB,

    E⃗0⋅r^B=0\vec E_0\cdot \hat r_B=0E0​⋅r^B​=0

    so

    E⃗B=3(E⃗0⋅r^B)r^B=0\vec E_B=3(\vec E_0\cdot \hat r_B)\hat r_B=0EB​=3(E0​⋅r^B​)r^B​=0

    Hence Option B is correct.

  1. Check option A

    Magnitude of total field is

    E=3∣E⃗0⋅r^∣=3E0∣cos⁡θ∣E = 3|\vec E_0\cdot \hat r| = 3E_0|\cos\theta|E=3∣E0​⋅r^∣=3E0​∣cosθ∣

    where θ\thetaθ is angle between r^\hat rr^ and E⃗0\vec E_0E0​.

    This clearly changes from point to point on the circle. Therefore the magnitude is not same at any two points.

    So Option A is false.

  1. Check option D

    Point AAA lies along the direction of p⃗\vec pp​ (and E⃗0\vec E_0E0​). Therefore at AAA,

    r^A∥E⃗0\hat r_A \parallel \vec E_0r^A​∥E0​

    so

    E⃗A=3E0r^A=3E02(i^+j^)\vec E_A = 3E_0\hat r_A = \frac{3E_0}{\sqrt2}(\hat i+\hat j)EA​=3E0​r^A​=2​3E0​​(i^+j^​)

    But option D states

    E⃗A=2E0(i^+j^)\vec E_A = \sqrt2 E_0(\hat i+\hat j)EA​=2​E0​(i^+j^​)

    whose magnitude is 2E02E_02E0​, not 3E03E_03E0​.

    Therefore Option D is false.

  1. Final answer

    Correct options are:

    B, C\boxed{B,\ C}B, C​

    This matches the stored correct answer.

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