91影视

Since the total charge q is uniformly distributed throughout the sphere, we can calculate the volume charge density 蟻 by dividing the total charge by the volume of the sphere. The volume of the sphere can be calculated using the formula \(V = \frac{4}{3}\pi R^3\) , so the volume charge density is given by: \(\rho = \frac{q}{\frac{4}{3}\pi R^3}\) .

For r 鈮� R, we will take a Gaussian surface as a sphere with radius r. The total enclosed charge on this Gaussian surface is equal to the total charge of the sphere which is q. According to Gauss's law, we can write: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}\) , where E is the electric field, and \(\epsilon_0\) is the permittivity of free space. The electric field is radial and symmetrical; thus, it is constant at any point on the Gaussian surface. Therefore, we can simplify the equation as: \(E \oint dA = \frac{q}{\epsilon_0}\) . Since the integral \(\oint dA\) represents the total surface area of the Gaussian sphere, we can write: \(E(4 \pi r^2) = \frac{q}{\epsilon_0}\) . From this equation, we can find the electric field outside the sphere (r 鈮� R) as: \(E(r) = \frac{q}{4 \pi \epsilon_0 r^2}\) , for r 鈮� R.

For r 鈮� R, we will again take a Gaussian surface as a sphere with radius r, but this time with r 鈮� R. The enclosed charge within this sphere can be determined by multiplying the volume charge density 蟻 by the volume enclosed by the Gaussian surface, which is \(\frac{4}{3}\pi r^3\) . Thus, we have: \(Q_{\text{enclosed}} = \rho \left(\frac{4}{3}\pi r^3\right) = \frac{q}{\frac{4}{3}\pi R^3} \left(\frac{4}{3}\pi r^3\right)\) . Applying Gauss's law, we can write: \(E \oint dA = \frac{Q_{\text{enclosed}}}{\epsilon_0}\) . Since the electric field is constant on the surface, we can simplify the equation as: \(E(4 \pi r^2) = \frac{q}{\epsilon_0} \frac{r^3}{R^3}\) . From this equation, we can find the electric field inside the sphere (r 鈮� R) as: \(E(r) = \frac{q}{4 \pi \epsilon_0 R^3} r\) , for r 鈮� R.

To sketch the graph, we will plot the two equations found in steps 2 and 3 for r 鈮� R and r 鈮� R, respectively. We know that for r 鈮� R, the electric field increases linearly with r, and for r 鈮� R, the electric field decreases with the square of the distance (1/r^2). The graph will show an increasing linear behavior for r 鈮� R and a decreasing curve for r 鈮� R, consistently meeting at the point (R, E(R)). This point represents the maximum electric field value on the surface of the sphere.