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\[{{180}^{\circ }}\].You need to login to perform this action.
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