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  • Can someone help me with this, - Electrostatics - NEET-4

The electric potential at a point (x, y, z) is given by V = -x2y - xz3 + 4. The electric field \vec{E} at that point is

  • Option 1)

    \vec{E}=\hat{i}2xy + \hat{j}(x^{2}+y^{2})+\hat{k}(3xz-y^{2})

  • Option 2)

    \vec{E}=\hat{i}z^{3}+\hat{j}xyz+\hat{k}z^{2}

  • Option 3)

    \vec{E}=\hat{i}(2xy-z^{3})+\hat{j}xy^{2}+\hat{k}3z^{2}x

  • Option 4)

    \vec{E}=\hat{i}(2xy+z^{3})+\hat{j}x^{2}+\hat{k}3zx^{2}

 

Answers (1)

 

When dielectric insert between the charges -

F_{med}=frac{F_{air}}{K}=frac{1q_{1}q_{2}}{4pi varepsilon _{0}kr^{2}}

- wherein

 

 \vec{E} =\vec{V}v =\frac{\vartheta v}{\vartheta y}\hat{j}-\frac{\vartheta v}{\vartheta y}\hat{j}-\frac{\vartheta v}{\vartheta z} \hat{k}

= -\left [ 2xy-z^{3} \right ]\hat{j}-\left [ -x^{2} \right ]\hat{j}-\left [ -3x^{2}z \right ]\hat{k}

\vec{E}=\left ( 2xy+z^{3} \right )\hat{j}+x^{2}\hat{j}+3x^{2}z\:\hat{k}


Option 1)

\vec{E}=\hat{i}2xy + \hat{j}(x^{2}+y^{2})+\hat{k}(3xz-y^{2})

Incorrect

Option 2)

\vec{E}=\hat{i}z^{3}+\hat{j}xyz+\hat{k}z^{2}

Incorrect

Option 3)

\vec{E}=\hat{i}(2xy-z^{3})+\hat{j}xy^{2}+\hat{k}3z^{2}x

Incorrect

Option 4)

\vec{E}=\hat{i}(2xy+z^{3})+\hat{j}x^{2}+\hat{k}3zx^{2}

Correct

Posted by

subam

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