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  • Need clarity, kindly explain! - Electrostatics - JEE Main-12

The electric potential V is given as a function of distance x (metre) by V=(5x^{2}+10x-9)volt . Value of electric field at x = 1m is         

  • Option 1)

    – 20 V/m

  • Option 2)

    6 V/m

  • Option 3)

    11 V/m

  • Option 4)

    – 23 V/m

 

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};  E=-\frac{d}{dx}(5x^{2}+10x-9)=(10x+10), at

x = 1m E=-20V/m 


Option 1)

– 20 V/m

Option 2)

6 V/m

Option 3)

11 V/m

Option 4)

– 23 V/m

Posted by

Vakul

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