91影视

Choose the Gaussian surface as a concentric sphere with radius r, where r is the distance from the center of the sphere. Using symmetry, it is evident that the electric field will have only a radial component.

Let's consider the spherical Gaussian surface with radius r inside the sphere (r < a). As the electric field (E) and the area vector (dA) are both radially pointing outward (same direction), the angle between them (胃) is 0 degrees. Therefore, the dot product of E and dA in Gauss's Law is |E||dA|cos胃 which simplifies to |E||dA|. To apply Gauss's Law, we need to determine the charge enclosed by the Gaussian surface of radius r. Since the sphere has a uniform charge density, we can find the charge density (蟻) as: 蟻 = Q / (4/3*蟺*a^3) The volume enclosed by the Gaussian surface is: V = (4/3)*蟺*r^3 Hence, the enclosed charge (q) is: q = 蟻*V = Q*(r^3) / (a^3) Now, apply Gauss's Law to the Gaussian surface of radius r: 鈭瓻鈰卍A = q / 蔚鈧� |E|鈭甦A = q / 蔚鈧� |E|(4蟺r^2) = (Q*r^3)/(a^3*蔚鈧�) Solving for E, we get for r < a: E = (Q*r)/(4蟺*蔚鈧�*a^3)

Now let's consider the spherical Gaussian surface with radius r outside the sphere (r > a). For this case, the angle (胃) between E and dA is still 0. The Gaussian surface encloses the whole sphere, hence enclosed charge (q) remains Q. Applying Gauss's Law to the Gaussian surface of radius r: 鈭瓻鈰卍A = q / 蔚鈧� |E|鈭甦A = Q / 蔚鈧� |E|(4蟺r^2) = Q/蔚鈧� Solving for E, we get for r > a: E = Q / (4蟺*蔚鈧�*r^2)

Since the electric field has only a radial component, we can write it in spherical coordinates as: E(r) = Er * r藛 = |E| * r藛 Combining our results for r < a and r > a, we get: Electric field E(r) = - For r < a: (Q*r) / (4蟺*蔚鈧�*a^3) * r藛 - For r > a: Q / (4蟺*蔚鈧�*r^2) * r藛