. 10 −9 ) 2 4π 0  R CA |R CA | 3 + R DA |R DA | 3 + R BA |R BA | 3  where R CA = a x −a y , R DA = a x + a y , and R BA =2a x . The magnitudes are |R CA | = |R DA | = √ 2, and |R BA | =2. Substituting these leads to F = (50 × 10 −9 ) 2 4π 0  1 2 √ 2 + 1 2 √ 2 + 2 8  a x =21.5a x µN where. a is a unit normal vector to the surface (as a look ahead (Chapter 4), note that taking the gradient of f gives a). 1.7. Given the vector field E =4zy 2 cos 2xa x +2zy sin 2xa y + y 2 sin 2xa z for. negative. This apparent sign ambiguity is not the real problem, however, as we really want the magnitude of the angle anyway. Choosing the positive sign, we are left with sin θ =7 √ 5/( √ 29 √ 9)=0.969.