Physics · 2021
JEE Main · 4 August 2021, Shift 2 · Q9
The electric potential at any point P ( x, y, z ) is V = x^3 z - x^2 y -3 volts. The electric field E_Q at point Q (2,3,1) will be (in V / m ) -
The electric potential at any point $\displaystyle \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is $\displaystyle \mathrm{V}=\mathrm{x}^3 \mathrm{z}-\mathrm{x}^2 \mathrm{y}-3$ volts.
The electric field $\displaystyle \overrightarrow{\mathrm{E}_{\mathrm{Q}}}$ at point $\displaystyle \mathrm{Q}(2,3,1)$ will be (in $\displaystyle \mathrm{V} / \mathrm{m}$ ) -
Official answer
(3)
$\displaystyle 4(\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.