Question

if electric field is given (5i+4j+9k) the electric flux through a surface of 20 units lying in X Y plane​

Answer 1

The electric field,

$\longrightarrow\vec{\sf{E}}=\sf{5\ \hat i+4\ \^j+9\ \^k}$

The surface of area 20 units is lying in XY plane so its areal vector is,

$\longrightarrow\vec{\sf{A}}=\sf{20\ \^k}$

We know the electric flux through a surface is equal to the dot product of electric field and area of the surface.

$\longrightarrow\Phi=\vec{\sf{E}}\cdot\vec{\sf{A}}$

Then here, electric flux through the surface is,

$\longrightarrow\Phi=\sf{\left(5\ \hat i+4\ \^j+9\ \^k\right)\cdot\left(20\ \^k\right)}$

$\longrightarrow\Phi=\sf{\left(5\ \hat i+4\ \^j+9\ \^k\right)\cdot\left(0\ \hat i+0\ \^j+20\ \^k\right)}$

$\longrightarrow\Phi=\sf{5\cdot 0+4\cdot 0+9\cdot20}$

$\longrightarrow\underline{\underline{\Phi=\sf{180}}}$

Hence 180 units is the answer.