A particle having a charge of \(1.6 \times 10^{-19} \mathrm{C}\) enters between the plates of a parallel plate capacitor. The initial velocity of the particle is parallel to the plates. A potential difference of \(300 \mathrm{v}\) is applied to the capacitor plates. If the length of the capacitor plates is $10 \mathrm{~cm}\( and they are separated by \)2 \mathrm{~cm}$, Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is \(12 \times 10^{-24} \mathrm{~kg}\). (A) \(10^{4}(\mathrm{~m} / \mathrm{s})\) (B) \(10^{2}(\mathrm{~m} / \mathrm{s})\) (C) \(10^{-1}(\mathrm{~m} / \mathrm{s})\) (D) \(10^{3}(\mathrm{~m} / \mathrm{s})\)

We can use the formula for the electric field inside a parallel plate capacitor: \[ E = \frac{V}{d} \] Where: - \( E \) is the electric field (in \( N/C \)) - \( V \) is the potential difference between the plates (in V) - \( d \) is the distance between the plates (in m) Given that \( V = 300 \, V \) and \( d = 2 \, cm = 0.02 \, m \), we can calculate the electric field \( E \): \[ E = \frac{300 \, V}{0.02 \, m} = 15\,000 \, N/C \]

Using the formula for electric force: \[ F_e = qE \] Where: - \( F_e \) is the electric force (in N) - \( q \) is the charge of the particle (in C) - \( E \) is the electric field (in \( N/C \)) Given that \( q = 1.6 \times 10^{-19} \, C \) and \( E = 15\,000 \, N/C \), we can calculate the electric force \( F_e \): \[ F_e = (1.6 \times 10^{-19} \, C)(15\,000 \, N/C) = 2.4 \times 10^{-15} \, N \]

Using the formula for acceleration: \[ a = \frac{F_e}{m} \] Where: - \( a \) is the acceleration (in \( m/s^2 \)) - \( F_e \) is the electric force (in N) - \( m \) is the mass of the particle (in kg) Given that \( F_e = 2.4 \times 10^{-15} \, N \) and \( m = 12 \times 10^{-24} \, kg \), we can calculate the acceleration \( a \): \[ a = \frac{2.4 \times 10^{-15} \, N}{12 \times 10^{-24} \, kg} = 2 \times 10^{8} \, m/s^2 \]

We can use the formula for the motion of the particle to calculate the time it takes to reach the top plate: \[ d = \frac{1}{2}at^2 \] Where: - \( d \) is the distance between the plates (in m) - \( a \) is the acceleration (in \( m/s^2 \)) - \( t \) is the time (in s) Given that \( d = 0.02 \, m \) and \( a = 2 \times 10^{8} \, m/s^2 \), we can solve for \( t \): \[ 0.02 \, m = \frac{1}{2}(2 \times 10^{8} \, m/s^2)t^2 \] Solving for \( t \): \[ t = \sqrt{\frac{0.02 \, m}{10^8 \, m/s^2}} = 10^{-4} \, s \] Now we need to calculate the time required to travel the plate length. We can use the formula: \[ x = vt \] Where: - \( x \) is the length of the plate (in m) - \( v \) is the initial velocity of the particle (in \( m/s \)) - \( t \) is the time (in s) Given that \( x = 10 \, cm = 0.1 \, m \) and \( t = 10^{-4} \, s \), we can now solve for \( v \) \[ 0.1 \, m = v(10^{-4} \, s) \] Solving for \( v \): \[ v = 10^4 \, m/s \] The greatest initial velocity for which the particle will not be able to come out of the plates is \( 10^4 \, m/s \). The correct answer is (A) \(10^{4}(\mathrm{~m} / \mathrm{s})\).