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If the electric potential at any point \mathrm{(x, y, z) m} in space is given by \mathrm{V=3 x^{2}} volt.

The electric field at the point \mathrm{(1,0,3) m} will be:

Option: 1

3 \, \mathrm{Vm}^{-1} \text {, directed along positive } x \text {-axis. }


Option: 2

3 \, \mathrm{Vm}^{-1} \text {, directed along negative } x \text {-axis. }


Option: 3

6 \, \mathrm{Vm}^{-1} \text {, directed along positive } x \text {-axis. }


Option: 4

6 \, \mathrm{Vm}^{-1} \text {, directed along negative } x \text {-axis. }


Answers (1)

best_answer

\mathrm{ V=3 x^{2}}\\

\mathrm{E=\frac{-d v}{d x}=-6 x}\\

\vec{E}=-6 x \hat{\imath}\\

\mathrm{\text { at }(1,0,3)}\\

\vec{E}=-6 \hat{\imath}

Hence correct option is 4

Posted by

Gautam harsolia

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