Question

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) \(10^{3}(\mathrm{~m} / \mathrm{s})\)

Short answer

The greatest initial velocity for which the particle will not be able to come out of the plates is (A) \(10^{4}(\mathrm{m}/\mathrm{s})\).

Step-by-step solution

Calculate the electric field inside the capacitor

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 \( E \): \[ E = \frac{300 \, V}{0.02 \, m} = 15\,000 \, N/C \]

Calculate the electric force acting on the particle

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 \( F_e \): \[ F_e = (1.6 \times 10^{-19} \, C)(15\,000 \, N/C) = 2.4 \times 10^{-15} \, N \]

Calculate the acceleration of the particle

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 \( a \): \[ a = \frac{2.4 \times 10^{-15} \, N}{12 \times 10^{-24} \, kg} = 2 \times 10^{8} \, m/s^2 \]

Calculate the time to reach the top plate and compare it with the time to travel the plate length.

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) \(10^{4}(\mathrm{~m} / \mathrm{s})\).