A) \[4.4\times {{10}^{6}}N-m,\text{ }3.2\times {{10}^{-4}}J\]
B) \[-2\times {{10}^{-3}}N-m,\text{ }-4\times {{10}^{3}}J\]
C) \[4\times {{10}^{3}},\text{ }2\times {{10}^{-3}}J\]
D) \[2\times {{10}^{-3}}N-m,\text{ }4\times {{10}^{-3}}J\]
Correct Answer: D
Solution :
\[q=\pm 1\times {{10}^{-6}}C\] \[2a=2.0cm=2.0\times {{10}^{-2}}m\] \[W=?,\,{{\theta }_{1}}={{0}^{o}},\,{{\theta }_{2}}={{180}^{o}}\] \[{{\tau }_{\max }}=pE=q(2a)E\] \[=1\times {{10}^{-6}}\times 2.0\times {{10}^{-2}}\times 1\times {{10}^{5}}\] \[=2\times {{10}^{-3}}Nm\] \[W=pE(\cos {{\theta }_{1}}-\cos {{\theta }_{2}})\] \[=({{10}^{-6}}\times 2\times {{10}^{-2}})({{10}^{5}})(\cos {{0}^{o}}-\cos {{180}^{o}})\] \[=4\times {{10}^{-3}}\,J\]You need to login to perform this action.
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