A) \[-1.9\times {{10}^{25}}J\]
B) zero
C) \[9.5\times {{10}^{-26}}J\]
D) \[1.9\times {{10}^{-25}}J\]
Correct Answer: D
Solution :
: Work done\[=pE(1-\cos \theta )\] \[=(6.2\times {{10}^{-30}})\times (1.5\times {{10}^{4}})(1-cos180{}^\circ )\] \[=2\times 6.2\times 1.5\times {{10}^{-26}};1.86\times {{10}^{-25}}J\] \[=1.9\times {{10}^{-25}}.\]You need to login to perform this action.
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