If potential (in volts) in a region is expressed as V(x, y, z) = 6xy - y + 2yz, the electric field (in N/C) at point (1, 1, 0) is :

  • Option 1)

    - \left( {6\hat i + 5\hat j + 2\hat k} \right)

  • Option 2)

    - \left( {2\hat i + 3\hat j + \hat k} \right)

  • Option 3)

    - \left( {6\hat i + 9\hat j + \hat k} \right)

  • Option 4)

    - \left( {3\hat i + 5\hat j + 3\hat k} \right)

 

Answers (1)

As we learnt in 

In space -

E_{x}=frac{-dv}{dx}  ,  E_{y}=frac{-dv}{dy}    ,  E_{z}=frac{-dv}{dz}

-

 

 

Relation between field and potential -

E=frac{-dv}{dr}

- wherein

frac{dv}{dr} -   Potential gradient.

 

 

E = -\left [ \frac{dv}{dx}\hat{j}+\frac{dv}{dy}\hat{j}+\frac{dv}{dz}\hat{k} \right ]

v(x,y,z) = 6xy-y+2yz

\therefore E =- \left [ \frac{d}{dx}\left ( 6xy-y+2yz \right )\hat{i}+\frac{d}{dy} \left ( 6xy-y+2yz \right )\hat{j} + \frac{d}{dz} \left ( 6xy-y+2yz \right )\hat{k}\right ]

= -\left [ \left ( 6y \right )\hat{i}+\left ( 6x-1+2z \right )\hat{j}+(2\left y \right ) \hat{k}\right ]

4t point \left ( 1, 1, 0 \right )

\vec{E} =-\left [ 6\left ( 1 \right )\hat{i}+6\left ( 1 \right )-1+2\left ( 0 \right ) \hat{j} +2\left ( 1 \right )\hat{k}\right ]

=\left -( 6\hat{i} +5\hat{j}+2\hat{k}\right )


Option 1)

- \left( {6\hat i + 5\hat j + 2\hat k} \right)

Correct

Option 2)

- \left( {2\hat i + 3\hat j + \hat k} \right)

Incorrect

Option 3)

- \left( {6\hat i + 9\hat j + \hat k} \right)

Incorrect

Option 4)

- \left( {3\hat i + 5\hat j + 3\hat k} \right)

Incorrect

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