To solve the problem, we need to find the minimum velocity of a charged bullet so that it can emerge from the other side of a uniformly charged dielectric sphere. Here are the steps to derive the solution:
### Step 1: Understand the Problem
We have a dielectric sphere with a radius \( R = 10 \, \text{cm} = 0.1 \, \text{m} \) and a total charge \( Q = 3 \, \text{mC} = 3 \times 10^{-3} \, \text{C} \). A bullet with charge \( q = 3 \, \text{mC} = 3 \times 10^{-3} \, \text{C} \) and mass \( m = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} \) is fired from the surface of the sphere towards its center.
### Step 2: Calculate the Electric Field Inside the Sphere
Using Gauss's law, the electric field \( E \) inside the sphere at a distance \( r \) from the center is given by:
\[
E = \frac{Q \cdot r}{4 \pi \epsilon_0 R^3}
\]
where \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \).
### Step 3: Calculate the Force on the Bullet
The force \( F \) experienced by the bullet due to the electric field is:
\[
F = q \cdot E = q \cdot \frac{Q \cdot r}{4 \pi \epsilon_0 R^3}
\]
### Step 4: Work Done Against the Electric Force
The work done \( W \) against the electric force as the bullet moves from the surface to the center of the sphere is given by:
\[
W = \int_{R}^{0} F \, dr = \int_{R}^{0} \left( q \cdot \frac{Q \cdot r}{4 \pi \epsilon_0 R^3} \right) dr
\]
Substituting the expression for \( F \):
\[
W = \frac{qQ}{4 \pi \epsilon_0 R^3} \int_{R}^{0} r \, dr
\]
Calculating the integral:
\[
\int r \, dr = \frac{r^2}{2} \Big|_{R}^{0} = 0 - \frac{R^2}{2} = -\frac{R^2}{2}
\]
Thus,
\[
W = -\frac{qQ}{4 \pi \epsilon_0 R^3} \cdot \left(-\frac{R^2}{2}\right) = \frac{qQ R^2}{8 \pi \epsilon_0 R^3} = \frac{qQ}{8 \pi \epsilon_0 R}
\]
### Step 5: Set the Work Done Equal to Kinetic Energy
The initial kinetic energy \( KE \) of the bullet is given by:
\[
KE = \frac{1}{2} m v^2
\]
Setting the work done equal to the kinetic energy:
\[
\frac{1}{2} m v^2 = \frac{qQ}{8 \pi \epsilon_0 R}
\]
### Step 6: Solve for Velocity \( v \)
Rearranging the equation to solve for \( v^2 \):
\[
v^2 = \frac{qQ}{4 \pi \epsilon_0 m R}
\]
Taking the square root gives:
\[
v = \sqrt{\frac{qQ}{4 \pi \epsilon_0 m R}}
\]
### Step 7: Substitute the Values
Substituting the known values:
- \( q = 3 \times 10^{-3} \, \text{C} \)
- \( Q = 3 \times 10^{-3} \, \text{C} \)
- \( m = 10 \times 10^{-3} \, \text{kg} \)
- \( R = 0.1 \, \text{m} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \)
Calculating \( v \):
\[
v = \sqrt{\frac{(3 \times 10^{-3})(3 \times 10^{-3})}{4 \pi (8.85 \times 10^{-12})(10 \times 10^{-3})(0.1)}}
\]
Calculating the value gives:
\[
v \approx 9 \times 10^{3} \, \text{m/s}
\]
### Final Answer
The minimum velocity of the bullet so that it can emerge from the other side of the sphere is approximately:
\[
v \approx 9000 \, \text{m/s}
\]
