Question

the potential in the air is given by the function 5xyz+6(square of x). calculate the charge density

Answer 1

Answer:

तुम्हारी बहन यू ही ऊपर

Answer 2

Given:

The potential in the air is given by the function $\phi=5xyz+6x^{2}$.

To Find:

Calculate the charge density?

Solution:

Potential in air is given by function $\phi=5xyz+6x^{2}$.

so, electric field is given by

$\[\begin{gathered} \vec E = - \vec \nabla \phi = - \left( {\hat i\frac{\partial }{{dx}} + \hat j\frac{\partial }{{dy}} + \hat k\frac{\partial }{{dz}}} \right)\left( {5xyz + 6{x^2}} \right) \hfill \\ \vec E = - \left( {\left( {5yz + 12x} \right)\hat i + (5xz)\hat j + (5xy)\hat k} \right) \hfill \\ \end{gathered} \]$

By, Gauss Law,

$\[\begin{gathered} E.4\pi {r^2} = \frac{{{q_{in}}}}{{{\varepsilon _0}}} \hfill \\ \Rightarrow {q_{in}} = \left( {\left( {5yz + 12x} \right)\hat i + (5xz)\hat j + (5xy)\hat k} \right)4\pi {r^2}{\varepsilon _0} \hfill \\ \end{gathered} \]$

Now,

$\dfrac{{d{{\vec q}_{in}}}}{{d\vec r}} = \dfrac{{\left( { - \left( {5yz + 12x} \right)\hat i + (5xz)\hat j + (5xy)\hat k} \right)4\pi {r^2}{\varepsilon _0}}}{{d(\vec x + \vec y + \vec z)}} \hfill$

and

$\vec V = \dfrac{4}{3}\pi ({x^3} + {y^3} + {z^3});\dfrac{{d\vec V}}{{d\vec r}} = \dfrac{{4\pi ({x^3} + {y^3} + {z^3})}}{{d(\vec x + \vec y + \vec z)}} \hfill$

Hence, charge density is given by-

$\[\dfrac{{d{{\vec q}_{in}}}}{{d\vec r}} \times \dfrac{{d\vec r}}{{d\vec V}} = \dfrac{{\left( { - \left( {5yz + 12x} \right)\hat i + (5xz)\hat j + (5xy)\hat k} \right)4\pi {r^2}{\varepsilon _0}}}{{d(\vec x + \vec y + \vec z)}} \times \dfrac{{d(\vec x + \vec y + \vec z)}}{{4\pi ({x^3} + {y^3} + {z^3})}}\]$