The electrons in a particle beam each have a kinetic energy of `1.6 xx 10^(-17)J`. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0cm

Options

A. `10^(3)` v/m in the direction of velocity of electrons
B. `10^(3)` v/m positive direction of velocity of electrons
C. `10^(3) v//m` perpendicular to velocity of electrons
D. `10^(6) v//m` perpendicular to velocity of electrons

Answer

To solve the problem, we need to determine the magnitude and direction of the electric field that stops the electrons in a distance of 10.0 cm, given that their kinetic energy is \(1.6 \times 10^{-17} \, \text{J}\).

### Step-by-Step Solution:

**Step 1: Understand the given information**
- The kinetic energy (KE) of the electrons is given as:
\[ KE = 1.6 \times 10^{-17} \, \text{J} \]
- The distance (d) over which the electrons are stopped is:
\[ d = 10.0 \, \text{cm} = 0.1 \, \text{m} \]
- The charge of an electron (q) is:
\[ q = -1.6 \times 10^{-19} \, \text{C} \]

**Step 2: Set up the work-energy principle**
- The work done (W) by the electric field on the electrons is equal to the change in kinetic energy. Since the electrons are stopped, their final kinetic energy is 0:
\[ W = KE_{\text{final}} - KE_{\text{initial}} = 0 - 1.6 \times 10^{-17} = -1.6 \times 10^{-17} \, \text{J} \]

**Step 3: Relate work done to electric field**
- The work done by the electric field can also be expressed as:
\[ W = F \cdot d = q \cdot E \cdot d \cdot \cos(\theta) \]
Here, \(F\) is the force, \(E\) is the electric field, and \(\theta\) is the angle between the force and the displacement. For this case, we can assume \(\theta = 0\) degrees (force and displacement are in the same direction), so \(\cos(0) = 1\):
\[ W = q \cdot E \cdot d \]

**Step 4: Substitute the known values**
- Since the charge of the electron is negative, we can write:
\[ -1.6 \times 10^{-17} = (-1.6 \times 10^{-19}) \cdot E \cdot 0.1 \]

**Step 5: Solve for the electric field (E)**
- Rearranging the equation gives:
\[ E = \frac{-1.6 \times 10^{-17}}{-1.6 \times 10^{-19} \cdot 0.1} \]
- Simplifying this:
\[ E = \frac{1.6 \times 10^{-17}}{1.6 \times 10^{-20}} = \frac{1.6}{1.6} \times 10^{3} = 10^{3} \, \text{N/C} \]

**Step 6: Determine the direction of the electric field**
- Since the electrons are negatively charged, they will experience a force in the direction opposite to the electric field. Therefore, the electric field must be directed opposite to the motion of the electrons to stop them.

### Final Answer:
- **Magnitude of the electric field**: \(E = 10^{3} \, \text{N/C}\)
- **Direction of the electric field**: Opposite to the direction of motion of the electrons.