A) \[6\sqrt{5}N\]
B) \[30\,N\]
C) \[24\,N\]
D) \[4\sqrt{35}N\]
Correct Answer: D
Solution :
[d] 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 \[(1,\text{ }1,\text{ }1)\] 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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