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Electrostatics question

2020 · 9 Jan · Shift 1 · Q50
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Electrostatics question

2020 · 9 Jan · Shift 1 · Q50

JEE MainPhysicsElectrostaticsMCQ+4 / −1
An electric dipole of moment p→=(−i^−3j^+2k^)×10−29\overrightarrow p = \left( { - \widehat i - 3\widehat j + 2\widehat k} \right) \times {10^{ - 29}}p​=(−i−3j​+2k)×10−29 C.m is at the origin (0, 0, 0). The electric field due to this dipole at r→=+i^+3j^+5k^\overrightarrow r = + \widehat i + 3\widehat j + 5\widehat kr=+i+3j​+5k(note that r→.p→=0\overrightarrow r .\overrightarrow p = 0r.p​=0 ) is parallel to :
  1. A
    (+i^+3j^−2k^)\left( { + \widehat i + 3\widehat j - 2\widehat k} \right)(+i+3j​−2k)
  2. B
    (+i^−3j^−2k^)\left( { + \widehat i - 3\widehat j - 2\widehat k} \right)(+i−3j​−2k)
  3. C
    (−i^+3j^−2k^)\left( { - \widehat i + 3\widehat j - 2\widehat k} \right)(−i+3j​−2k)
  4. D
    (−i^−3j^+2k^)\left( { - \widehat i - 3\widehat j + 2\widehat k} \right)(−i−3j​+2k)
View written solutionFree

Correct answer: A

  1. Electric field due to a dipole

For a dipole moment p⃗\vec pp​ at position vector r⃗\vec rr, the electric field is

E⃗=14πε0r3[3(p⃗⋅r^)r^−p⃗].\vec E = \frac{1}{4\pi\varepsilon_0 r^3}\left[3(\vec p\cdot \hat r)\hat r - \vec p\right].E=4πε0​r31​[3(p​⋅r^)r^−p​].

Equivalently,

E⃗=14πε0[3(p⃗⋅r⃗)r⃗r5−p⃗r3].\vec E = \frac{1}{4\pi\varepsilon_0}\left[\frac{3(\vec p\cdot \vec r)\vec r}{r^5} - \frac{\vec p}{r^3}\right].E=4πε0​1​[r53(p​⋅r)r​−r3p​​].
  1. Given vectors
p⃗=(−i^−3j^+2k^)×10−29 C⋅m\vec p = (-\hat i - 3\hat j + 2\hat k)\times 10^{-29}\ \text{C·m}p​=(−i^−3j^​+2k^)×10−29 C⋅m

and

r⃗=i^+3j^+5k^.\vec r = \hat i + 3\hat j + 5\hat k.r=i^+3j^​+5k^.
  1. Check the dot product
r⃗⋅p⃗∝(1)(−1)+(3)(−3)+(5)(2)=−1−9+10=0.\vec r\cdot \vec p \propto (1)(-1) + (3)(-3) + (5)(2) = -1 - 9 + 10 = 0.r⋅p​∝(1)(−1)+(3)(−3)+(5)(2)=−1−9+10=0.

So indeed,

p⃗⋅r⃗=0.\vec p\cdot \vec r = 0.p​⋅r=0.

Hence the first term in the dipole field formula vanishes:

E⃗=−14πε0p⃗r3.\vec E = -\frac{1}{4\pi\varepsilon_0}\frac{\vec p}{r^3}.E=−4πε0​1​r3p​​.

Therefore, E⃗\vec EE is parallel to −p⃗-\vec p−p​.

  1. Find the direction of −p⃗-\vec p−p​

Since

p⃗∥(−i^−3j^+2k^),\vec p \parallel (-\hat i - 3\hat j + 2\hat k),p​∥(−i^−3j^​+2k^),

we get

−p⃗∥(i^+3j^−2k^).-\vec p \parallel (\hat i + 3\hat j - 2\hat k).−p​∥(i^+3j^​−2k^).
  1. Match with options

This corresponds to:

(i^+3j^−2k^)\boxed{(\hat i + 3\hat j - 2\hat k)}(i^+3j^​−2k^)​

which is Option A.

  1. Verification of other options
  • A: (i^+3j^−2k^)(\hat i + 3\hat j - 2\hat k)(i^+3j^​−2k^) ✅ matches −p⃗-\vec p−p​
  • B: (i^−3j^−2k^)(\hat i - 3\hat j - 2\hat k)(i^−3j^​−2k^) ❌ wrong jjj-component
  • C: (−i^+3j^−2k^)(-\hat i + 3\hat j - 2\hat k)(−i^+3j^​−2k^) ❌ wrong iii-component
  • D: (−i^−3j^+2k^)(-\hat i - 3\hat j + 2\hat k)(−i^−3j^​+2k^) ❌ this is parallel to p⃗\vec pp​, not −p⃗-\vec p−p​

So the correct answer is A.

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