A) \[-10\times {{10}^{-29}}J\]
B) \[-7\times {{10}^{-27}}J\]
C) \[-20\times {{10}^{-18}}J\]
D) \[-9\times {{10}^{-20}}J\]
Correct Answer: B
Solution :
\[E=1000V/m,p={{10}^{-29}}cm,\theta ={{45}^{o}}\] Potential energy stored in the dipole, \[U=-\vec{p}.\vec{E}=-pE\cos \theta \] \[=-{{10}^{-29}}\times 1000\times \cos {{45}^{o}}=-\frac{1}{\sqrt{2}}\times {{10}^{-26}}\] \[=-0.707\times {{10}^{-26}}J\approx -7\times {{10}^{-27}}J\]
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