
An electric dipole moment of dipole is $\vec{p}=\left( -\hat{i}-3\hat{j}+2\hat{k} \right)\times {{10}^{-29}}C.m$ is at the origin $\left( 0,0,0 \right)$. The electric field due to this dipole at $\hat{r}=\left( \hat{i}+3\hat{j}+5\hat{k} \right)$ (note that $\vec{r}\cdot \vec{p}=0$) parallel is to:
A. $\left( -\hat{i}+3\hat{j}-2\hat{k} \right)$
B. $\left( -\hat{i}-3\hat{j}+3\hat{k} \right)$
C. $\left( \hat{i}+3\hat{j}-2\hat{k} \right)$
D. $\left( \hat{i}-3\hat{j}-2\hat{k} \right)$
Answer
501.3k+ views
Hint: The electric dipole moment is the product of the magnitude of the charge and the distance between the two charges. This is a useful concept for the study of dielectrics and other solid applications. The electric field of a dipole is always in the opposite direction (antiparallel) to that of the dipole moment.
As per the given data,
The electric dipole moment at origin is $\vec{p}=\left( -\hat{i}-3\hat{j}+2\hat{k} \right)\times {{10}^{-29}}C.m$
The position vector is given as $\hat{r}=\left( \hat{i}+3\hat{j}+5\hat{k} \right)$
The dot product of electric dipole moment and position vector is zero ($\vec{r}\cdot \vec{p}=0$)
Complete answer:
When two charges are kept parted by a certain distance it is known as a dipole. The product of the magnitude of the charge and the distance between the two charges is termed as an electric dipole moment.
Mathematically,
$p=qd$
Where,
$q$ is the magnitude of the charge
$d$ is the distance between two charges
The dot product is used to find the magnitude of the resultant quantity. Mathematically it is calculated as,
\[a\cdot b=\left| a \right|\left| b \right|\cos \theta \]
As it is mentioned in the question that the dot product of the electric dipole moment and the position vector is zero ($\vec{r}\cdot \vec{p}=0$).
So this can be written as,
$\begin{align}
& \vec{r}\cdot \vec{p}=\left| r \right|\left| p \right|\cos \theta =0 \\
& \Rightarrow \cos \theta =0 \\
& \Rightarrow \theta =\dfrac{\pi }{2} \\
\end{align}$
So here we can say that the electric field and position vector are perpendicular to each other.
The electric field of a dipole is given as,
$\vec{E}=-\lambda p$
Here the minus sign represents that the electric field is in the opposite (anti-parallel) direction to that of the electric dipole moment.
For the situation mention in the question the electric field of the dipole can be given as,
$\begin{align}
& \vec{E}=-\vec{p} \\
& \Rightarrow \vec{E}=-\left( -\hat{i}-3\hat{j}+2\hat{k} \right) \\
& \therefore \vec{E}=\left( \hat{i}+\hat{j}-2\hat{k} \right) \\
\end{align}$
Thus, from the above discussion, the correct option which satisfies the given question is Option C.
Note:
The concept of electric dipole moment is also useful in atoms and molecules where the effects of charge separation can be measured easily. We can also find the potential of a dipole at a distance point using the superposing the point charge potentials. The potential at the point decreases with an increase in the distance between the point and the dipole.
As per the given data,
The electric dipole moment at origin is $\vec{p}=\left( -\hat{i}-3\hat{j}+2\hat{k} \right)\times {{10}^{-29}}C.m$
The position vector is given as $\hat{r}=\left( \hat{i}+3\hat{j}+5\hat{k} \right)$
The dot product of electric dipole moment and position vector is zero ($\vec{r}\cdot \vec{p}=0$)
Complete answer:
When two charges are kept parted by a certain distance it is known as a dipole. The product of the magnitude of the charge and the distance between the two charges is termed as an electric dipole moment.
Mathematically,
$p=qd$
Where,
$q$ is the magnitude of the charge
$d$ is the distance between two charges
The dot product is used to find the magnitude of the resultant quantity. Mathematically it is calculated as,
\[a\cdot b=\left| a \right|\left| b \right|\cos \theta \]
As it is mentioned in the question that the dot product of the electric dipole moment and the position vector is zero ($\vec{r}\cdot \vec{p}=0$).
So this can be written as,
$\begin{align}
& \vec{r}\cdot \vec{p}=\left| r \right|\left| p \right|\cos \theta =0 \\
& \Rightarrow \cos \theta =0 \\
& \Rightarrow \theta =\dfrac{\pi }{2} \\
\end{align}$
So here we can say that the electric field and position vector are perpendicular to each other.
The electric field of a dipole is given as,
$\vec{E}=-\lambda p$
Here the minus sign represents that the electric field is in the opposite (anti-parallel) direction to that of the electric dipole moment.
For the situation mention in the question the electric field of the dipole can be given as,
$\begin{align}
& \vec{E}=-\vec{p} \\
& \Rightarrow \vec{E}=-\left( -\hat{i}-3\hat{j}+2\hat{k} \right) \\
& \therefore \vec{E}=\left( \hat{i}+\hat{j}-2\hat{k} \right) \\
\end{align}$
Thus, from the above discussion, the correct option which satisfies the given question is Option C.
Note:
The concept of electric dipole moment is also useful in atoms and molecules where the effects of charge separation can be measured easily. We can also find the potential of a dipole at a distance point using the superposing the point charge potentials. The potential at the point decreases with an increase in the distance between the point and the dipole.
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