B) Using Gauss's theorem, calculate the flux of the vector field F = x

1.	B) Using Gauss's theorem, calculate the flux of the vector field F = xi+yi+zkthrough the surface of a cylinder of radius A and height H, which has its axisalong the z-axis. The base of the cylinder is on the xy plane.
Answer» The flux of given VECTOR FIELD is Φ =3πA²H STEP by step explanation $\textbf{\Large According to gauss theorem :}\\\text{\large Total electric flux by a closed surface in electric field is equal to = }4\pi \Sigma q\\\text{where ,total charge elclosed in closed surface =}\Sigma q\\\\\text{by gauss theorem\\}\\ \phi = \oint_{\partial V} \left( \boldsymbol{F} \cdot \boldsymbol{n} \right) dS = \int_V \left( \nabla \cdot \boldsymbol{F} \right) d V\\\\Here , \\\nabla \cdot \boldsymbol{F} = \textbf{Divergence of vector field is = 1 + 1 + 1 =} 3$ because $\nabla \cdot \boldsymbol{F}= \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z}$ $\textbf{Putting this value in gauss theorem}$ the equation becomes $\phi = \int_V \left( \nabla \cdot \boldsymbol{F} \right) d V = \int_V \left( 3\right) d V = 3V\\\\\textbf{\Large the volume of this cylinder =} \pi A^2 H\\\\So \\\textbf{\Large The flux of the vector field is = } \phi = 3\times (\pi A^2 H) = 3\pi A^2H$

B) Using Gauss's theorem, calculate the flux of the vector field F = xi+yi+zkthrough the surface of a cylinder of radius A and height H, which has its axisalong the z-axis. The base of the cylinder is on the xy plane.

Answer»

The flux of given VECTOR FIELD is Φ =3πA²H

STEP by step explanation

$\textbf{\Large According to gauss theorem :}\\\text{\large Total electric flux by a closed surface in electric field is equal to = }4\pi \Sigma q\\\text{where ,total charge elclosed in closed surface =}\Sigma q\\\\\text{by gauss theorem\\}\\ \phi = \oint_{\partial V} \left( \boldsymbol{F} \cdot \boldsymbol{n} \right) dS = \int_V \left( \nabla \cdot \boldsymbol{F} \right) d V\\\\Here , \\\nabla \cdot \boldsymbol{F} = \textbf{Divergence of vector field is = 1 + 1 + 1 =} 3$ because

$\nabla \cdot \boldsymbol{F}= \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z}$

$\textbf{Putting this value in gauss theorem}$

the equation becomes

$\phi = \int_V \left( \nabla \cdot \boldsymbol{F} \right) d V = \int_V \left( 3\right) d V = 3V\\\\\textbf{\Large the volume of this cylinder =} \pi A^2 H\\\\So \\\textbf{\Large The flux of the vector field is = } \phi = 3\times (\pi A^2 H) = 3\pi A^2H$