A) 8 along negative \[X-\]axis
B) 8 along positive \[X-\]axis
C) 16 along negative \[X-\]axis
D) 16 along positive \[Z-\]axis
Correct Answer: A
Solution :
The electric potential \[V\,(x,y,z)=4{{x}^{2}}\,volt\] Now \[\overrightarrow{E}=-\,\left( \hat{i}\frac{\partial V}{\partial x}+\hat{j}\frac{\partial V}{\partial y}+\hat{k}\frac{\partial V}{\partial z} \right)\] Now \[\frac{\partial V}{\partial x}=8x,\,\frac{\partial V}{\partial y}=0\] and \[\frac{\partial V}{\partial z}=0\] Hence \[\overrightarrow{E}=-\,8x\hat{i}\], so at point (1m, 0, 2m) \[\overrightarrow{E}=-\,8\hat{i}\,\,volt/metre\] or 8 along negative X-axis.
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