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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Mass of the particle, $m = 100 \,\text{mg} = 1 \times 10^{-4} \,\text{kg}$
• Charge on the particle, $q = 40\,\mu\text{C} = 40 \times 10^{-6}\,\text{C}$
• Electric field strength, $E = 1 \times 10^{5}\,\text{N\,C}^{-1}$
• Initial velocity, $u = 200\,\text{m\,s}^{-1}$ (opposite to the direction of the electric field)
• Final velocity (momentarily at rest), $v = 0\,\text{m\,s}^{-1}$
Step 2: Write the Kinematic Equation
We use the standard equation of motion for constant acceleration:
$ v^2 - u^2 = 2\,a\,s $
Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $s$ is the distance traveled.
Step 3: Determine the Acceleration from the Electric Force
The force on the charged particle due to the electric field is given by:
$ F = qE $
According to Newton’s second law, $F = ma$, so the acceleration is:
$ a = \frac{F}{m} = \frac{qE}{m} $
Since the particle is thrown opposite to the direction of the electric field and the electric force acts along the field, the acceleration will be negative relative to the particle’s initial velocity.
Step 4: Substitute Acceleration into the Kinematic Equation
Using $v^2 - u^2 = 2as$ and $v = 0$,
$ 0^2 - (200)^2 = 2 \left(-\,\frac{qE}{m}\right) s $
The negative sign indicates that the acceleration is opposite to the initial velocity. Rearrange to solve for $s$:
$ - (200)^2 = 2 \Bigl(-\,\frac{(40 \times 10^{-6}) \times (1 \times 10^{5})}{1 \times 10^{-4}} \Bigr) s $
Step 5: Calculate the Distance
First, compute the term for acceleration:
$ qE = 40 \times 10^{-6} \times 10^{5} = 40 \times 10^{-1} = 4 \,\text{N} \\
a = \frac{4}{1 \times 10^{-4}} = 4 \times 10^{4} \,\text{m\,s}^{-2} = 40000\,\text{m\,s}^{-2}
$
Substitute $a = -40000\,\text{m\,s}^{-2}$ into the equation:
$ -(200)^2 = 2 \times \bigl(-40000\bigr)\,s \\
-40000 = -80000\,s \\
s = \frac{-40000}{-80000} = 0.5\,\text{m}
$
Final Answer: $0.5\,\text{m}$
