A charged particle with charge $q$ and mass $m$ is given an initial kinetic energy $K_0$ at the middle of a uniformly charged spherical region of total charge $Q$ and radius $R$. $q$ and $Q$ have opposite signs. The spherically charged region is not free to move. Throughout this problem consider electrostatic forces only.

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(a) Find the value of $K_0$ such that the particle will just reach the boundary of the spherically charged region.

(b) How much time does it take for the particle to reach the boundary of the region if it starts with the kinetic energy $K_0$ found in part (a)?
### Part (a)

1. **Using Gauss's Law to find the electric field inside the sphere:**

   Consider a Gaussian surface of radius \( r \) inside the uniformly charged sphere. The charge enclosed by this Gaussian surface is given by:
   \[
   Q_{\text{enc}} = \frac{r^3}{R^3} Q
   \]
   By Gauss's Law:
   \[
   \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
   \]
   Since the electric field \( E(r) \) is radially symmetric:
   \[
   E(r) \cdot 4\pi r^2 = \frac{r^3 Q}{R^3 \epsilon_0}
   \]
   Solving for \( E(r) \):
   \[
   E(r) = \frac{Qr}{4\pi R^3 \epsilon_0}
   \]

2. **Finding the potential energy difference from the center to the boundary:**

   The potential energy \( V \) at a distance \( r \) from the center is given by:
   \[
   V(r) = - \int_0^r E(r') \, dr'
   \]
   Substituting \( E(r') \):
   \[
   V(r) = - \int_0^r \frac{Qr'}{4\pi R^3 \epsilon_0} \, dr'
   \]
   Evaluating the integral:
   \[
   V(r) = - \frac{Q}{4\pi R^3 \epsilon_0} \int_0^r r' \, dr' = - \frac{Q}{4\pi R^3 \epsilon_0} \left[ \frac{r'^2}{2} \right]_0^r = - \frac{Q r^2}{8\pi R^3 \epsilon_0}
   \]
   At the boundary \( r = R \):
   \[
   V(R) = - \frac{Q R^2}{8\pi R^3 \epsilon_0} = - \frac{Q}{8\pi R \epsilon_0}
   \]

3. **Determining the initial kinetic energy \( K_0 \):**

   The initial kinetic energy \( K_0 \) must equal the potential energy difference to just reach the boundary:
   \[
   K_0 = \frac{-qQ}{8\pi R \epsilon_0}
   \]

### Part (b)

1. **Finding the force on the particle:**

   The force \( F(r) \) on the particle at a distance \( r \) from the center is:
   \[
   F(r) = q E(r) = q \cdot \frac{Qr}{4\pi R^3 \epsilon_0} = \frac{qQr}{4\pi R^3 \epsilon_0}
   \]
   This force is analogous to the restoring force of a spring, \( F = -kr \), where:
   \[
   k = \frac{qQ}{4\pi R^3 \epsilon_0}
   \]

2. **Determining the time to reach the boundary:**

   The motion of the particle is simple harmonic motion (SHM) with angular frequency \( \omega \):
   \[
   \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{qQ}{4\pi R^3 \epsilon_0 m}}
   \]
   The time to reach the boundary (quarter of the period of SHM) is:
   \[
   T = \frac{2\pi}{\omega}
   \]
   Therefore, the time to reach the boundary is:
   \[
   t = \frac{T}{4} = \frac{\pi}{2\omega} = \frac{\pi}{2} \sqrt{\frac{4\pi R^3 \epsilon_0 m}{qQ}}
   \]

The final answer is \( \boxed{ K_0 = \frac{-qQ}{8\pi R \epsilon_0} } \) and \( t = \frac{\pi}{2} \sqrt{\frac{4\pi R^3 \epsilon_0 m}{qQ}} \)