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  • Electric potential is given by volt. The electric field at point <img alt="\mathrm{

Electric potential is given by \mathrm{v=3 x^2} volt. The electric field at point \mathrm{(2,0,4) \mathrm{m}} will be -

Option: 1

\mathrm{12 \mathrm{Vm}^{-1} } directed along negative \mathrm{ \mathrm{x} }- axes. 


Option: 2

10 \mathrm{Vm}^{-1} directed along negative \mathrm{x} - axes. 


Option: 3

12 \mathrm{vm}^{-1} directed along positive \mathrm{x}- axes. 


Option: 4

10 \mathrm{vm}^{-1} directed along positive  \mathrm{y}- axes. 


Answers (1)

best_answer

\mathrm{\begin{aligned} \vec{E}_x & =-\frac{d v}{d x} \hat{\imath} \\ & =-\frac{d}{d x}\left[3 x^2\right] \hat{\imath} \end{aligned}}
\mathrm{\begin{gathered} \vec{E}_x=-6 x \hat{i} \\ \text { at point }(2,0,4) \Rightarrow \vec{E}_x=-12 \hat{i} \mathrm{~V} / \mathrm{m} . \end{gathered}}

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