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An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work in rotating it through the angle of ${{180}^{0}}$.

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Last updated date: 17th Jul 2024
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Answer
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Hint: The first part of the problem can be solved by using the formula for the force on a charge in an electric field. A dipole has two charges of equal magnitude and opposite signs. The second part of the problem can also be solved by using the direct formula for the external work required to rotate a dipole in an electric field between two angles (that the dipole axis makes with the direction of the electric field).

Formula used:
$\overrightarrow{F}=q\overrightarrow{E}$
$W=pE\left( \cos {{\theta }_{1}}-\cos {{\theta }_{2}} \right)$

Complete step by step answer:
(i) A dipole consists of two charges of equal magnitude but opposite signs. When a dipole is placed in an electric field, the total force on it is the sum of the force on each charge due to the electric field.
The force $\overrightarrow{F}$ on a charge $q$ placed in an electric field $\overrightarrow{E}$ is given by
$\overrightarrow{F}=q\overrightarrow{E}$ --(1)
We will consider a dipole with charges $+q$ and $-q$ placed in an electric field $\overrightarrow{E}$.
Now, using (1), we get the force $\overrightarrow{{{F}_{p}}}$ on charge $+q$ due to the electric field as
$\overrightarrow{{{F}_{p}}}=+q\overrightarrow{E}$ --(2)
Similarly, using (1), we get the force $\overrightarrow{{{F}_{n}}}$ on charge $-q$ due to the electric field as
$\overrightarrow{{{F}_{n}}}=-q\overrightarrow{E}$ --(3)
As explained above, the total force on the dipole due to the electric field will be the sum of the forces on the two charges. Hence, we get the total force $\overrightarrow{F}$ on the dipole placed in the electric field as
$\overrightarrow{F}=\overrightarrow{{{F}_{p}}}+\overrightarrow{{{F}_{n}}}$
Using (2) and (3), we get,
$\overrightarrow{F}=+q\overrightarrow{E}+\left( -q\overrightarrow{E} \right)=+q\overrightarrow{E}-q\overrightarrow{E}=0$
Hence, the net force on a dipole placed in an electric field is zero.

(ii) Now, when a dipole is placed in an electric field, external work is required to rotate it between two angles. The angles refer to the angle made by the dipole axis with the direction of the electric field.
The external work $W$ required to rotate a dipole of dipole moment of magnitude $p$ in an electric field of magnitude $E$ is given by
$W=pE\left( \cos {{\theta }_{1}}-\cos {{\theta }_{2}} \right)$ --(4)
where ${{\theta }_{1}},{{\theta }_{2}}$ are the initial and final angles made by the dipole axis with the direction of the electric field.
Now, we will consider an electric dipole of dipole moment magnitude $p$ placed in an electric field of magnitude $E$.
Let ${{\theta }_{1}},{{\theta }_{2}}$ be the initial and final angles made by the dipole axis with the direction of the electric field.
It is given that the dipole is initially aligned along the electric field.
$\therefore {{\theta }_{1}}={{0}^{0}}$
Also, it is given that the dipole has to be rotated through ${{180}^{0}}$.
$\therefore {{\theta }_{2}}={{0}^{0}}+{{180}^{0}}={{180}^{0}}$
Using (4), the external work $W$ required to rotate the dipole will be
$W=pE\left( \cos \left( {{0}^{0}} \right)-\cos \left( {{180}^{0}} \right) \right)=pE\left( 1-\left( -1 \right) \right)=pE\left( 1+1 \right)=pE\left( 2 \right)=2pE$
$\left( \because \cos \left( {{0}^{0}} \right)=1,\cos \left( {{180}^{0}} \right)=-1 \right)$
Therefore, we have found the amount of external work required.

Note: The position where the dipole axis is parallel to the electric field is the position of stable equilibrium for the dipole. In this position, it has the lowest potential energy and hence, when a slight deflection is given to the dipole, it will try to come back to this position. On the other hand, when the dipole axis is anti-parallel to the electric field direction, the potential energy of the dipole is the highest and it is in a position of unstable equilibrium. If the dipole is given a slight deflection at this point it will keep on rotating to reach the point of stable equilibrium. Students must remember this important concept.