Verified Solution • 2025–26 Session moderate

QJEEM20Ph826c: JEE Main 2020 Physics - Class 12 JEEM20Ph826c Solution: An electric dipole of moment C.m is at the origin (0, 0, 0).…

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}} 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 k (note that r→.p→=0\overrightarrow r .\overrightarrow p = 0 ) is parallel to :

Multiple Choice Options

  • A (+i^+3j^−2k^)\left( { + \widehat i + 3\widehat j - 2\widehat k} \right)
  • B (+i^−3j^−2k^)\left( { + \widehat i - 3\widehat j - 2\widehat k} \right)
  • C (−i^+3j^−2k^)\left( { - \widehat i + 3\widehat j - 2\widehat k} \right)
  • D (−i^−3j^+2k^)\left( { - \widehat i - 3\widehat j + 2\widehat k} \right)

Step-by-Step Verified Solution

Since r→\overrightarrow r and p→\overrightarrow p are perpendicular to each other therefore point lies on the equitorial plane. Therefore electric field at the point will be antiparallel to the dipole moment.

∴\therefore E→∣∣−p→\overrightarrow E || - \overrightarrow p

⇒\Rightarrow E→∣∣(i^+3j^−2k^)\overrightarrow E ||\left( {\widehat i + 3\widehat j - 2\widehat k} \right)

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