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  • Need clarity, kindly explain! - Electrostatics - JEE Main-11

The electric potential V at any point x, y, z (all in metres) in space is given by  V=4x^{2}volt.The electric field at the point (1m, 0, 2m) in volt/metre is

  • Option 1)

    8 along negative X-axis

  • Option 2)

    8 along positive X-axis

  • Option 3)

    16 along negative X-axis

  • Option 4)

    16 along positive Z-axis

 

Answers (1)

As we learned

 

Relation between field and potential -

E=\frac{-dv}{dr}

- wherein

\frac{dv}{dr} -   Potential gradient.

 

 By using  E=-\frac{dV}{dx}\Rightarrow E=-\frac{d}{dx}(4x^{2})=-8x. Hence at point (1m, 0, 2m).  E=-8 volt/mi.e. 8 along – vex-axis.


Option 1)

8 along negative X-axis

Option 2)

8 along positive X-axis

Option 3)

16 along negative X-axis

Option 4)

16 along positive Z-axis

Posted by

Vakul

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