question_answer

A dipole moment of a dipole is 16 x 10 26 mC and is situated in an electric Field of intensity 104 V/m, its minimum energy will be:                                                                                                                  [BHU M-2012]

A)                  \[3.18\times {{10}^{8}}\,m/s\]                 

B)                  \[2.12\times {{10}^{8}}\,m/s\]

C)                  \[2.90\times {{10}^{8}}\,m/s\]                 

D)                  \[0.\text{18 gauss},\text{ }0.\text{18}\sqrt{3}\text{ gauss}\]

Correct Answer: B

Solution :

                 Key Idea: When dipole is rotated through \[{{90}^{\circ }}\] from the direction of the field, them minimum energy is obtained.                 The work done in retaining an electric dipole in uniform electric field through an angle\[\theta \], from the direction of field is                 \[W=-pE\left( 1-\cos \,\theta  \right)\]                                 This work is stored as potential energy. When\[\theta ={{90}^{\circ }}\], potential energy is minimum.                 Therefore, \[W=-pE\]                 Given, \[p=16\times {{10}^{-26}}\,mC\]                                 \[E={{10}^{4}}\,V/m\]                                 \[W=-16\times {{10}^{-26}}\times {{10}^{4}}\]                                 \[W=-16\times {{10}^{-22}}\,\,J\]                 Note: Maximum energy is obtained when dipole is rotated through\[{{180}^{\circ }}\].