The electric potential at a point ( {\x}, {\y}, {\z}) is given by V=-x^2 y-x z^3+4 The electric field \overrightarrow{ {\E} at that point is :

Correct Answer is : →E=∧i(2xy+z3)+∧jx2+∧k3xz2

\overrightarrow{ {\E}={ {\i^{∧}}}(2 x y- {\z}^3)+{ {\j^{∧}} x y}{ }^2+{ {\k^{∧}}} 3 {\z}^2 {\x}

\overrightarrow{ {\E}={ {\i^{∧}}}(2 x y+ {\z}^3)+{ {\j^{∧}}} {\x}^2+{ {\k^{∧}}} 3 x z^2

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

\overrightarrow{ {\E}={ {\i^{∧}}} z+{ {\j^{∧}}} {\xyz}+{ {\k^{∧}}} {\z}^2

AIPMT 2009 Physics and Chemistry Paper

Question : 30 of 100

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The electric potential at a point (x,y,z) is given by V=−x2y−xz3+4 The electric field

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at that point is :

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