An electron enters in high potential region `V_(2)` from lower potential region `V_(1)` then its velocity

Options

A. will increase
B. will change in direction but not in magnitude
C. will not change in direction of field
D. will no change in direction perpendicular to field

Answer

To solve the problem of how the velocity of an electron changes when it moves from a lower potential region \( V_1 \) to a higher potential region \( V_2 \), we can use the principle of conservation of mechanical energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Conservation of Mechanical Energy
The mechanical energy of the electron is conserved as it moves from one potential region to another. This means that the sum of its kinetic energy (KE) and potential energy (PE) remains constant.

### Step 2: Write the Energy Conservation Equation
The conservation of mechanical energy can be expressed as:
\[ KE_1 + PE_1 = KE_2 + PE_2 \]
Where:
- \( KE_1 \) is the kinetic energy at potential \( V_1 \)
- \( PE_1 \) is the potential energy at potential \( V_1 \)
- \( KE_2 \) is the kinetic energy at potential \( V_2 \)
- \( PE_2 \) is the potential energy at potential \( V_2 \)

### Step 3: Express Kinetic and Potential Energies
The kinetic energy of the electron is given by:
\[ KE = \frac{1}{2} mv^2 \]
The potential energy of a charge \( q \) in a potential \( V \) is given by:
\[ PE = qV \]
For an electron, the charge \( q = -e \) (where \( e \) is the elementary charge).

### Step 4: Substitute the Energies into the Equation
Substituting the expressions for kinetic and potential energy into the conservation equation, we get:
\[ \frac{1}{2} mv_1^2 - eV_1 = \frac{1}{2} mv_2^2 - eV_2 \]

### Step 5: Rearrange the Equation
Rearranging the equation to isolate the kinetic energies gives:
\[ \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2 = e(V_2 - V_1) \]

### Step 6: Analyze the Change in Kinetic Energy
Since \( V_2 > V_1 \) (the electron moves to a higher potential), the term \( e(V_2 - V_1) \) is positive. This indicates that the kinetic energy at \( V_2 \) is greater than at \( V_1 \):
\[ KE_2 > KE_1 \]

### Step 7: Relate Kinetic Energy to Velocity
Since kinetic energy is related to velocity, we can conclude that:
\[ \frac{1}{2} mv_2^2 > \frac{1}{2} mv_1^2 \]
This implies that the velocity \( v_2 \) must be greater than \( v_1 \):
\[ v_2 > v_1 \]

### Conclusion
Thus, when an electron enters a high potential region \( V_2 \) from a lower potential region \( V_1 \), its velocity increases.

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