A charge of mass .m. charge .2e. is released from rest in a uniform electric field of strength .E.. The time taken by it to travel a distance .d. in the field is

Options

A. `t= sqrt((dm)/(Ee))`
B. `sqrt((2dm)/(Ee))`
C. `sqrt((2dE)/(m e))`
D. `sqrt((2Ee)/(dm))`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given quantities
- Mass of the charge, \( m \)
- Charge, \( q = 2e \) (where \( e \) is the elementary charge)
- Electric field strength, \( E \)
- Distance traveled, \( d \)

### Step 2: Calculate the force acting on the charge
The force \( F \) on a charge in an electric field is given by the formula:
\[ F = qE \]
Substituting the value of charge:
\[ F = (2e)E = 2eE \]

### Step 3: Determine the acceleration of the charge
Using Newton's second law, \( F = ma \), we can find the acceleration \( a \):
\[ a = \frac{F}{m} = \frac{2eE}{m} \]

### Step 4: Use the equations of motion to find the time taken
Since the charge is released from rest, the initial velocity \( u = 0 \). We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting \( s = d \), \( u = 0 \), and \( a = \frac{2eE}{m} \):
\[ d = 0 + \frac{1}{2} \left(\frac{2eE}{m}\right) t^2 \]
This simplifies to:
\[ d = \frac{eE}{m} t^2 \]

### Step 5: Solve for time \( t \)
Rearranging the equation to solve for \( t^2 \):
\[ t^2 = \frac{dm}{eE} \]
Taking the square root of both sides gives:
\[ t = \sqrt{\frac{dm}{eE}} \]

### Final Answer
The time taken by the charge to travel a distance \( d \) in the electric field is:
\[ t = \sqrt{\frac{dm}{eE}} \]

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