Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The electric dipole of moment $\begin{array}{*{20}{c}} {\overrightarrow p }& = &{( - \widehat i - 3\widehat j + 2\widehat k) \times {{10}^{ - 29}}Cm} \end{array}$is at the origin (0,0,0). The electric field due to this dipole at $\begin{array}{*{20}{c}} {\overrightarrow r }& = &{(\widehat i + 3\widehat j + 5\widehat k)} \end{array}$is parallel to [Note that $\begin{array}{*{20}{c}} {\overrightarrow {r.} \overrightarrow p }& = &0 \end{array}$] A) $\widehat i - 3\widehat j - 2\widehat k$B) $- \widehat i - 3\widehat j + 2\widehat k$C ) $\widehat i + 3\widehat j - 2\widehat k$D) $- \widehat i + 3\widehat j - 2\widehat k$

Last updated date: 08th Sep 2024
Total views: 77.7k
Views today: 1.77k
Verified
77.7k+ views
Hint: In this question, we have given a condition:
($\begin{array}{*{20}{c}} {\overrightarrow {r.} \overrightarrow p }& = &0 \end{array}$),
Therefore, According to the given condition we will get to know that the $\overrightarrow p$and position vector $\overrightarrow r$will be perpendicular to each other. Due to the opposite direction of the electric field to the electric dipole moment, the electric field vector will be the same as the electric dipole moment but in a negative sign. Hence, we will get a suitable answer.

Basically, the direction of the electric dipole moment is from negative charge to positive charge while the direction of the electric field is from positive charge to negative charge. In other words, we can say that the direction of the electric field is opposite to the electric dipole moment. Hence, as per the given note $\begin{array}{*{20}{c}} {\overrightarrow {r.} \overrightarrow p }& = &0 \end{array}$, we can conclude that the electric dipole moment and electric field will be perpendicular to each other but will be in opposite directions.

Electric dipole moment vector and position vector are given,
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow p }& = &{( - \widehat i - 3\widehat j + 2\widehat k) \times {{10}^{ - 29}}Cm} \end{array}$ and position vector $\begin{array}{*{20}{c}} {\overrightarrow r }& = &{(\widehat i + 3\widehat j + 5\widehat k)} \end{array}$

Therefore, As per given the condition, we can write that
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow {r.} \overrightarrow p }& = &0 \end{array}$

The $\overrightarrow p$and $\overrightarrow r$are perpendicular to each other. It means that the electric field will also be perpendicular to the position vector $\overrightarrow r$.
$\begin{array}{*{20}{c}} { \Rightarrow (\widehat i + 3\widehat j + 5\widehat k).( - \widehat i - 3\widehat j + 2\widehat k)}& = &0 \end{array}$

Now, we know that the electric field is directly proportional to the electric dipole moment, but will be in the opposite direction. There will be a constant. Whose value will be greater than 0.
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow E }& = &{ - \lambda \overrightarrow p } \end{array}$

Here, a negative sign indicates that the electric field is in the opposite direction to the electric dipole moment. Therefore,
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow E }& = &{ - \lambda ( - \widehat i - 3\widehat j + 2\widehat k) \times {{10}^{ - 29}}} \end{array}$
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow E }& = &{\lambda (\widehat i + 3\widehat j - 2\widehat k) \times {{10}^{ - 29}}} \end{array}$
Therefore,$\overrightarrow E$will be parallel to the $(\widehat i + 3\widehat j - 2\widehat k)$

Now, the final answer is $(\widehat i + 3\widehat j - 2\widehat k)$. So, the correct option is C.

Note: In this question, the first point is to keep in mind that the $\begin{array}{*{20}{c}} \lambda & > &0 \end{array}$, Where, $\lambda$is constant which is the replacement of the proportionality sign.