Question

A circular surface with a radius of 0.072 misexposed to a uniform external electricfield of magnitude 1.44x10⁴ N/C. The electric flux through the surface is 82 Nm²/C. What is the angle between the direction of the electric field and the area vector?​

Answer 1

$A=\pi r^2=3.14\cdot0.072^2=0.0163\ (m^2)$

$\Phi=E\cdot A\cdot\cos\alpha\to\alpha=\cos^{-1}(\frac{\Phi}{E\cdot A})=$

$=\cos^{-1}(\frac{82}{1.44\cdot10^4\cdot 0.0163})\approx70°$ . Answer