In a region of uniform electric field ofn intencity E, an electron of mass `m_e` is released from rest. The distance travelled by the eloctron in a time t is

Options

A. `(2m_et^2)/e`
B. `(eEt^2)/(2m_e)`
C. `(m_e"gt"^2)/(eE)`
D. `(2Et^2)/(em_e)`

Answer

To solve the problem step-by-step, we will follow the principles of physics related to electric fields and motion.

### Step 1: Identify the forces acting on the electron
The electron is placed in a uniform electric field \( E \). The force \( F \) acting on the electron due to the electric field can be expressed as:
\[ F = qE \]
where \( q \) is the charge of the electron. The charge of the electron is \( -e \) (where \( e \) is the elementary charge). Thus, the force acting on the electron is:
\[ F = -eE \]

### Step 2: Relate force to acceleration
According to Newton's second law, the force acting on an object is also equal to the product of its mass and acceleration:
\[ F = m_e a \]
where \( m_e \) is the mass of the electron and \( a \) is its acceleration. Setting the two expressions for force equal gives:
\[ m_e a = -eE \]
From this, we can solve for the acceleration \( a \):
\[ a = -\frac{eE}{m_e} \]

### Step 3: Use kinematic equations to find the distance traveled
The electron is released from rest, which means its initial velocity \( u = 0 \). We can use the kinematic equation for distance \( s \) traveled under constant acceleration:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \) into the equation gives:
\[ s = \frac{1}{2} a t^2 \]
Now, substituting the expression for acceleration \( a \):
\[ s = \frac{1}{2} \left(-\frac{eE}{m_e}\right) t^2 \]
This simplifies to:
\[ s = -\frac{eE}{2m_e} t^2 \]

### Step 4: Interpret the result
The negative sign indicates that the electron moves in the direction opposite to the electric field due to its negative charge. However, for the distance traveled, we can express it as:
\[ s = \frac{eE}{2m_e} t^2 \]

### Final Answer
The distance traveled by the electron in time \( t \) is:
\[ s = \frac{eE}{2m_e} t^2 \]