A) \[6\sqrt{5}N\]
B) 30 N
C) 24 N
D) \[4\sqrt{35}N\]
Correct Answer: D
Solution :
We know \[F=qE\] ?(i) \[E=-\frac{dv}{dr}\] \[\left. \begin{align} & {{E}_{x}}\frac{\delta V}{\delta x}=6-8y \\ & {{E}_{y}}=\frac{\delta V}{\delta y}=-8x-8+6z \\ \end{align} \right\}\] ?(ii) \[{{E}_{z}}=6y\] Above values of \[{{E}_{x}},{{E}_{y}}\], and \[{{E}_{z}}\] at (1, 1, 1) are \[{{E}_{x}}=6-8\times (1)=-2\] \[{{E}_{y}}=-8(1)-8+6(1)=-10\] \[{{E}_{z}}=6\times 1=6\] So, \[{{E}_{net}}=\sqrt{{{(-2)}^{2}}+{{(10)}^{2}}+{{(6)}^{2}}}\] \[=\sqrt{4+100+36}=\sqrt{40}\Rightarrow \sqrt{35\times 4}\] \[=2\sqrt{35}N/C\] So, \[F=q\,{{E}_{net}}=2(2\sqrt{35})=4\sqrt{35}N\]
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