## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$106.02\times 10^{-2}Nm^{2}/C$
Given: $d=3cm$, so $r=1.5\times 10^{-2}m^{2}$ $A=\pi (1.5\times 10^{-2}m)^{2}j=2.25\pi\times 10^{-4}m^{2}j$ (plane xz) $E=1500i+1500j-1500k$ Electric flux $\phi$ through any surface is given as $\phi= E.A$ Thus,$\phi=(1500i+1500j-1500k)\times(2.25\pi\times 10^{-4}j)$ Hence, $\phi=1500\times2.25\pi\times 10^{-4}=106.02\times 10^{-2}Nm^{2}/C$