Strictly speaking, Coulomb`s law cannot be derived from Gauss`s law alone, since Gauss`s law does not provide information on the curvature of E (see Helmholtz decomposition and Faraday`s law). However, Coulomb`s law can be proved from Gauss`s law if it is further assumed that the electric field of a point charge is spherically symmetric (this assumption, like Coulomb`s law itself, is exactly true if the charge is stationary, and approximately true if the charge is moving). Gauss`s theorem also provides an important conclusion: since the perpendicular to the surface points is along the electric field, θ = 0. where r̂ is a unit vector pointing radially away from the charge. Again, by spherical symmetry, E points in the radial direction, and so we get Suppose we need to find the field at point P. All points on this surface are equivalent and by symmetry the field at all these points will be of the same size and radial in the direction. where E is the electric field, dA is a vector representing an infinitesimal surface element of the surface,[Note 2] and · Represents the point product of two vectors. Then we apply Φ=∫SE→·n^dAΦ=∫SE→·n^dA to this system and replace the known values. On the sphere n^=r^n^=r^ and r=Rr=R, i.e. for an infinitesimal surface dA, where the total area of the spherical surface is 4πR2.4πR2. This gives the flow through the spherical surface closed in radius r as But the total charge given to this hollow sphere is 6 × 10-8 C.
Therefore, the load on the outer surface is 10 × 10-8C. Thanks to the divergence theorem, Gauss`s law can alternatively be written in differential form: One of the fundamental relations between the two laws is that Gauss`s law can be used to derive Coulomb`s law and vice versa. We can further say that Coulomb`s law corresponds to Gauss`s law, which means that they are almost identical. Although this relationship is discussed in detail in electrodynamics, we consider a derivation using an example. To establish the relationship, let`s first take a look at Gaussian law. Consider a load of point q. If we now apply Coulomb`s law, the electric field generated is given by: Therefore, the net number of electric field lines passing from the inside to the outside through the two surfaces is equal. This net number of electric field lines, obtained by subtracting the number of lines in the outward direction from the number of lines in the inward direction, gives a visual measure of the electric flow through the surfaces. If the particle receives a charge q, the electric force qE acts upwards. It balances the weight of the particle if But because e ( r , r ′ ) ∈ C 1 ( V × Ω ) {displaystyle e(mathbf {r,mathbf {r} `} )in C^{1}(Vtimes Omega )} , Although microscopically all charges are basically the same, there are often practical reasons to want to treat related loads differently from free loads. It follows that the more fundamental Gaussian law with regard to E (above) is sometimes put in the equivalent form below, which is only in relation to D and gratuitousness. It is true for all r ≠ r ′ {displaystyle mathbf {r} neq mathbf {r`} } that r ⋅ e ( r , r ′ ) = 0 {displaystyle nabla _{mathbf {r} }cdot mathbf {e} (mathbf {r,r`} )=0} ∇.
Note that we only deal with differential forms, not integral forms, but this is sufficient, since differential and integral forms are equivalent, respectively, by the divergence theorem. The total flow of the electric field through the closed surface is therefore zero. According to Gauss`s law, the total load in the closed surface must be zero. The charge on the inner surface of A must be the same and opposite to that on the inner surface of B. Problem 5: A particle of mass 5 × 10-6g is maintained on a large layer of horizontal charge of density 4.0 × 10-6 C/m2 (figure). What charge should be given to this particle so that when it is released, it does not fall? How many electrons need to be removed to release this charge? How much mass is reduced by removing these electrons? These equations can be used to redraw the distribution shown in Figure (a, b) as shown in the Figure. Φ = → E. d → A = ∫E. dA cos 0 + ∫E.
dA cos 90° + ∫E. dA cos 90° Consider an infinitely long load line, where the load per unit length is λ. We can take advantage of the cylindrical symmetry of this situation. By symmetry, the electric fields all point radially away from the charging line, there is no component parallel to the charging line. The final equality follows by observing ( Ω ∖ B R ( r 0 ) ) ∩ B R ( r 0 ) = ∅ {displaystyle (Omega setminus B_{R}(mathbf {r} _{0}))cap B_{R}(mathbf {r} _{0})=emptyset } , and the argument above. The net flow of the surface on the left is different from zero because it contains a net charge. The net flow of the surface on the right is zero because there is no load. Task 6: Two printed circuit boards A and B are arranged parallel to each other. A receives a Q1 charge and B a Q2 charge.
Find the distribution of loads on the four surfaces. The net flow through a closed surface is directly proportional to the net load in the volume bounded by the closed surface. (Q1 – q)/2Aε0 – q/2Aε0 + q/2Aε0 – (Q2 + q)/2Aε0 Another way to see why the flow through a closed spherical surface is independent of the radius of the surface is to look at electric field lines. Note that any field line of q that pierces the surface in radius R1R1 also pierces the surface at R2R2 (Figure 6.14). The charge of an electron is 1.6 × 10-19C. The number of electrons to be removed; Now consider r 0 ∈ Ω {displaystyle mathbf {r} _{0}in Omega } and B R ( r 0 ) ⊆ Ω {displaystyle B_{R}(mathbf {r} _{0})subseteq Omega } as the sphere centered in r 0 {displaystyle mathbf {r} _{0}} with R {displaystyle R} as radius (it exists because Ω {displaystyle Omega } is an open set). for each volume V. For this equation to hold simultaneously for any possible volume V, it is necessary (and sufficient) that the integers are the same everywhere. Therefore, this equation is equivalent to: Since the point P is inside the conductor, this field must be zero. Due to radial symmetry, the curved surface is equidistant from the charging line and the electric field in the surface has a constant size throughout. To effectively apply Gauss`s Law, you need to have a clear understanding of what each term represents in the equation. The E→E→ field is the total electric field at any point on the Gaussian surface.
This global field includes the contributions of charges both inside and outside the Gaussian surface. However, qencqenc is only the charge in the Gaussian surface. After all, the Gaussian surface is a surface closed in space. This surface can correspond to the actual surface of a conductor, or it can be an imaginary geometric surface. The only requirement for a Gaussian surface is that it be closed (Figure 6.18). 3. The strength of the electric field near a plane charge layer is E = σ/2ε0K, where σ = surface charge density. According to Gaussian law, the total load must be zero.
Therefore, the load on the inner surface of the hollow sphere is 4 × 10-8C.
