The electric potential at a point (x, y, z) is given by y=-x^ 2 y-x z^ 3 +4 . The…

Physics · JEE Main · NTA Exams — Electrostatics

The electric potential at a point (x, y, z) is given by \(y=-x^{2} y-x z^{3}+4\).

The electric field at that point is

\(\vec{E}=\hat{i}\left(z x y+z^{3}\right)+\hat{j} x^{2}+\hat{k} 3 x z^{2}\)
\(\vec{E}=\hat{i} 2 x y+\hat{j}\left(x^{2}+y^{2}\right)+\hat{k}\left(3 x z-y^{2}\right)\)
\(\vec{E}=\hat{i} z^{3}+\hat{j} x y z+\hat{k} z^{2}\)
\(\vec{E}=\hat{i}\left(2 x y-z^{3}\right)+\hat{j} x y^{2}+\hat{k} 3 z^{2} x\)

Answer

(A) E = i (z x y+z^ 3 )+ j x^ 2 + k 3 x z^ 2

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