A) \[0.9\text{ }N\]
B) \[0.1\text{ }N\]
C) \[3\times {{10}^{2}}N\]
D) \[0.6N\]
Correct Answer: D
Solution :
Maximum Electric field E = \[{{\vec{E}}_{0}}=(3\times {{10}^{-5}})c\left( -\hat{j} \right)\] \[{{\vec{E}}_{1}}=(2\times {{10}^{-6}})c\left( -\hat{i} \right)\] Maximum force \[{{\vec{F}}_{net}}=q\vec{E}=qc\left( -3\times {{10}^{-5}}\hat{j}-2\times {{10}^{-6}}\hat{i} \right)\] \[{{\vec{F}}_{0\max }}={{10}^{-4}}\times 3\times {{10}^{8}}\sqrt{{{(3\times {{10}^{-5}})}^{2}}+{{(2\times {{10}^{-6}})}^{2}}}\]\[=0.9N\] \[{{F}_{rms}}=\frac{{{F}_{0}}}{\sqrt{2}}=0.6N\] (approx) OptionYou need to login to perform this action.
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