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www.GOALias.blogspot.com(iv) The surface that we choose for the application of Gauss’s law is calledthe Gaussian surface. You may choose any Gaussian surface andapply Gauss’s law. However, take care not to let the Gaussian surfacepass through any discrete charge. This is because electric field dueto a system of discrete charges is not well defined at the location ofany charge. (As you go close to the charge, the field grows withoutany bound.) However, the Gaussian surface can pass through acontinuous charge distribution.(v) Gauss’s law is often useful towards a much easier calculation of theelectrostatic field when the system has some symmetry. This isfacilitated by the choice of a suitable Gaussian surface.(vi) Finally, Gauss’s law is based on the inverse square dependence ondistance contained in the Coulomb’s law. Any violation of Gauss’slaw will indicate departure from the inverse square law.Electric Chargesand FieldsExample 1.11 The electric field components in Fig. 1.27 areE x= αx 1/2 , E y= E z= 0, in which α = 800 N/C m 1/2 . Calculate (a) theflux through the cube, and (b) the charge within the cube. Assumethat a = 0.1 m.FIGURE 1.27Solution(a) Since the electric field has only an x component, for facesperpendicular to x direction, the angle between E and ΔS is± π/2. Therefore, the flux φ = E.ΔS is separately zero for each faceof the cube except the two shaded ones. Now the magnitude ofthe electric field at the left face isE L= αx 1/2 = αa 1/2(x = a at the left face).The magnitude of electric field at the right face isE R= α x 1/2 = α (2a) 1/2(x = 2a at the right face).The corresponding fluxes areφ L= E L.ΔS = ΔSE ˆL ⋅ nL=E LΔS cosθ = –E LΔS, since θ = 180°= –E La 2φ R= E R.ΔS = E RΔS cosθ = E RΔS, since θ = 0°= E Ra 2Net flux through the cubeEXAMPLE 1.1135