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Try this! - Electrostatics - JEE Main-4

3Electric potential at any point is  V=-5x+3y+\sqrt{15}z, then the magnitude of the electric field is

  • Option 1)

    3\sqrt{2}

     

     

  • Option 2)

    4\sqrt{2}

  • Option 3)

    5\sqrt{2}

  • Option 4)

    7

 
Answers (1)
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As we learned

 

Relation between E and V in integral form -

\dpi{100} dV=-\int_{r_{1}}^{r_{2}}\overrightarrow{E}\cdot \vec{d}r=-\int_{r_{1}}^{r_{2}}Edr\cos \theta

-

 

E_{x}=-\frac{dV}{dx}=-(-5)=5; E_{y}=-\frac{dV}{dy}=-3

and  E_{z}=-\frac{dV}{dz}=-\sqrt{15}

E_{net}=\sqrt{E_{x}^{2}+E_{y}^{2}+E_{z}^{2}}=\sqrt{(5)^{2}+(-3)^{2}+(-\sqrt{15})^{2}}=7 


Option 1)

3\sqrt{2}

 

 

Option 2)

4\sqrt{2}

Option 3)

5\sqrt{2}

Option 4)

7

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