A particle of charge $q = 1e$ and mass $m$ with kinetic energy $K$ enters an electric field set up by two parallel plates of length $l$ as illustrated in the figure. The potential difference between the two plates is $1\text{ V}$ and their separation is $d$. What is the minimum value of $K$ (in $\text{eV}$) for which the particle will not hit either of the plates? [$e$ is the charge of the electron.]}
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For a particle entering midway between parallel plates, the threshold for not hitting is when the exit deflection $y = d/2$.
Since $y \propto 1/K$, a higher kinetic energy leads to smaller deflection.
So setting $y = d/2$ gives the minimum value of $K$ directly.
Step 1: Understanding the Question:
This problem analyzes the projectile-like motion of a charged particle entering a uniform electric field between two parallel plates. We need to find the minimum kinetic energy required to prevent the particle from hitting the plates. Step 2: Key Formulas and Approach:
1. Electric field between plates:
\[ E = \frac{V}{d} \]
2. Force and acceleration:
\[ F = qE \implies a_y = \frac{qV}{md} \]
3. Kinematic equations of motion:
\[ x = vt \implies t = \frac{l}{v} \]
\[ y = \frac{1}{2} a_y t^2 \]
4. Relationship with kinetic energy:
\[ K = \frac{1}{2} mv^2 \implies mv^2 = 2K \] Step 3: Detailed Explanation:
The particle of mass $m$ and charge $q$ enters the plates exactly halfway between them. Therefore, the maximum allowable vertical deflection before hitting a plate is:
\[ y_{\max} = \frac{d}{2} \]
The horizontal velocity of the particle is $v$. The time $t$ taken to traverse the horizontal plate length $l$ is:
\[ t = \frac{l}{v} \]
The electric field $E$ exerts a constant vertical force $F = qE = \frac{qV}{d}$ on the particle, producing a vertical acceleration:
\[ a_y = \frac{qV}{md} \]
The vertical displacement $y$ of the particle as it leaves the plates is:
\[ y = \frac{1}{2} a_y t^2 = \frac{1}{2} \left( \frac{qV}{md} \right) \left( \frac{l}{v} \right)^2 = \frac{qVl^2}{2mv^2d} \]
Expressing the denominator in terms of kinetic energy $K = \frac{1}{2}mv^2 \implies mv^2 = 2K$:
\[ y = \frac{qVl^2}{4Kd} \]
To ensure the particle does not strike either plate, we must have $y<\frac{d}{2}$:
\[ \frac{qVl^2}{4Kd}<\frac{d}{2} \]
\[ \frac{qVl^2}{2Kd}<d \implies K>\frac{qVl^2}{2d^2} \]
Given that $q = 1e$ and $V = 1\text{ V}$, we substitute these values:
\[ K>\frac{(1e)(1\text{ V})l^2}{2d^2} = \left( \frac{l^2}{2d^2} \right) \text{eV} \]
Thus, the minimum kinetic energy in $\text{eV}$ is:
\[ K_{\min} = \frac{l^2}{2d^2} \]
Step 4: Final Answer:
The minimum kinetic energy is $\frac{l^2}{2d^2}\text{ eV}$, which corresponds to Option (A).
