Question

Consider the uniform electric field E =(4.0 ​j​^​​+3.0 ​k​^​​)×10​3^​​ N/C. What is its electric flux through a circular area of radius 1.34 m that lies in the xy-plane? Answer in Nxm^2/C

Answer 1

Electric flux through an area $\vec{S}$ due to an electric field $\vec{E}$ is $\Phi= \vec{E}\cdot\vec{S}$

Here electric field $\vec{E}=(4.0\hat{j}+3.0\hat{k})*10^{3}\,N/C$ and area vector is $\vec{S}=\pi R^2\hat{k}=(\pi *1.34^2)\hat{k}$

$\Phi= \vec{E}\cdot\vec{S}=(4.0\hat{j}+3.0\hat{k})*10^{3}\cdot(\pi *1.34^2)\hat{k}$

$\Phi=3.0*10^{3}*\pi *1.34^2=1.69*10^{4}\,N.m^2/C$