Suppose \(q\) is outside the irregularly shaped surface in Figure 24.26. Show that the electric flux due to \(q\) through the closed surface is zero. (Draw a small wedge from \(q\) through the surface and determine the electric flux through the two intersections between the wedge and the surface.)

Transcribed Image Text:

Figure 24.26 Formal proof of Gauss's law. (a) spherical Gaussian surface 1 spherical Gaussian surface 2, irregular surface wedge dA dA, dA (b) dA: intersection of wedge and spherical Gaussian surface 2 dA: intersection of wedge dA, and irregular surface dA: intersection of wedge and spherical Gaussian surface 1 dA di E