A) \[-\text{ }4\text{ }V{{m}^{-}}^{1}\]
B) \[+\,4\text{ }V{{m}^{-}}^{1}\]
C) \[-\,8\text{ }V{{m}^{-}}^{1}\]
D) \[8\text{ }V{{m}^{-}}^{1}\]
Correct Answer: C
Solution :
As, \[\operatorname{V}=2{{x}^{2}}\] \[\therefore \,\,\,\,\,\,\,{{E}_{x}}=\frac{-\partial V}{\partial x}=-\frac{\partial {{(2x)}^{2}}}{\partial x}=-\,4x\] \[{{E}_{Y}}=\,\,\frac{-\partial V}{\partial y}=0,\,\,{{E}_{Z}}=\frac{-\,\partial \,{{V}^{2}}}{\partial z}\,\,=\,\,0\] \[\therefore \,\,\,\,{{E}_{net}}={{E}_{x}}=-\,4x\] At, \[\left( 2m, 0,\,\,2m \right)\] \[{{E}_{net}}\,\,=\,\,(-\,4)\times 2\,\,=\,\,-8V{{m}^{-\,1}}\]
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