Question bank/JEE Main/Physics

An electric dipole of moment
$\overset{\rightarrow}{p} = ( - \overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 2\overset{\hat{}}{k}) \times 10^{- 29}$
cm. is at the origin (0,0,0). The electric fielddue to this dipole at
$\overset{\rightarrow}{r} = ( + \overset{\hat{}}{i} + 3\overset{\hat{}}{j} + 5\overset{\hat{}}{k})$
(note that$\overset{\rightarrow}{r}.\overset{\rightarrow}{p} = 0$) is
parallel to,d fo|qr f}/kzqo ftldk vk?kw.kZ ¼moment) [IMAGE] C.m.
[IMAGE] ij cuus okys fo|qr

A$( + \overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$B$( - \overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$C$( + \overset{\hat{}}{i} - 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$D$( - \overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 2\overset{\hat{}}{k})$

Show answer and solution▾

Since [IMAGE] must be antiparallel to [IMAGE], izfrlekUrj gksxk
So,[IMAGE] Where [IMAGE] is a arbitrary positive constant
Let
$\overset{\rightarrow}{A} = a\overset{\hat{}}{i} + b\overset{\hat{}}{j} + c\overset{\hat{}}{k}$
is parallel to $\overset{\rightarrow}{E}$
So, blfy, [IMAGE]

Source: Grass-G / P-21 Physcis Paper 19 Dec Hindi Eng..docx