91Ó°ÊÓ

For a uniformly charged sphere with total charge Q and radius R, the volume charge density \(\rho\) is given by: \(\rho = \frac{Q}{\frac{4}{3}\pi R^3}\) The electric field at distance r from the center is given by Gauss's Law. Let's apply Gauss's Law separately for inside and outside the sphere.

We can use Gauss's Law on a Gaussian surface with a spherical symmetry. Let's consider a Gaussian surface of radius r (r < R) inside the sphere. Gauss's Law states: \(E \cdot 4\pi r^2 = \frac{Q_{enclosed}}{\varepsilon_0}\) where, E is the electric field, \(\varepsilon_0\) is the vacuum permittivity, Q_enclosed is the charge enclosed by the Gaussian surface of radius r. For r < R, \(Q_{enclosed} = \rho \cdot \frac{4}{3}\pi r^3\) Now, replacing Q_enclosed in Gauss's Law equation from the above expression, we get: \(E \cdot 4\pi r^2 = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\varepsilon_0}\) Now, solve for E: \(E = \frac{\rho r}{3\varepsilon_0}\) Replace the value of \(\rho\) using the charge density equation. \(E = \frac{Qr}{4\pi \varepsilon_0 R^3}\) for r < R

Now let's consider a Gaussian surface of radius r (r ≥ R) outside the sphere. For r ≥ R, the charge enclosed by the Gaussian surface is the total charge Q of the sphere: \(E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0}\) Now, solve for E: \(E = \frac{Q}{4\pi \varepsilon_0 r^2}\) for r ≥ R

Now we have the electric field both inside and outside the sphere. The electric field function E(r) is: \(E(r) = \begin{cases} \frac{Qr}{4\pi \varepsilon_0 R^3} & (r < R) \\ \frac{Q}{4\pi \varepsilon_0 r^2} & (r \ge R) \end{cases}\) Now, plot the electric field E as a function of distance r from the center of the sphere. Inside the sphere (r < R), the electric field will increase linearly with r. Outside the sphere (r ≥ R), the electric field will decrease as the square of the distance r from the center.