The flux through an element of area delta A vector is phiE

The flux through an element of area delta A vector is: phi_E = E vector . delta A vector = | E | | delta A | cos(theta) where theta is the angle between E vector and delta A vector. Evaluate this for area delta A_1 vector, as: phi_E = (1214 N/C)(6.0 times 10^-5 m^2) cos(135 degree) = For area delta A_2 vector we obtain: phi_E = (1214 N/C)(6.0 times 10^-5 m^2) cos(90 degree) = And for area delta A vector_3: phi_E = (1214 N/C)(6.0 times 10^-5 m^2) cos(65 degree) = Find the orientation a surface increment must have to produce maximum flux.

Solution

2. phi = 1214*6*10^-5*cos 135 deg = 1214*6*10^-5*(-1/1.414)= - 5.15*10^-2 deg

4. phi = 1214*6*10^-5*cos 45 deg = 1214*6*10^-5*(1/1.414)= 5.15*10^-2 deg

3. phi = 0 as cos 90 = 0