Question

The electric field between two parallel plates is uniform, with magnitude 698 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.04 cm. At the same moment, both particles are released.

Required:
a. Calculate the distance (in cm) from the positive plate at which the two pass each other.
b. Repeat part (a) for a sodlum lon (Na+) and a chlorlde lon (CI-).

Answer 1

x = 4.03 10⁻² m

Step-by-step explanation:

Let's start by finding the acceleration for each particle due to the electric field

F = ma

the electric force is F = qE

q E = m a

a = qE / m

proton

m = 1.67 10⁻²⁷ kg

a₁ = 1.6 10⁻¹⁹ 698 /1.67 10⁻²⁷

a₁ = 6.687 10¹⁰ m / s²

directed to the right

electron

m = 9.11 10⁻³¹ kg

a₂ = 1.6 10⁻¹⁹ 698 /9.11 10⁻³¹

a₂ = 1.23 10¹⁴ m / s²

directed to the left

Taking the acceleration of the two bodies, we set a reference system with zero at the initial position of the proton on the positive plate, the point where it is located is x for the proton and x for the electron,

for the proton

x₁ = x₀₁ + v₀₁ t + ½ a₁ t²

as we start from rest vo1 = 0 and the initial position is xo = 0

x₁ = ½ a₁ t²

for the electron

x₂ = x₀₂ + v₀₂ t + ½ a₂ t²

in this case the initial velocity is zero v₀₂ = 0 and the initial position is x₀₂=d

x₂ = x₀₂ + ½ a₂ t²

at the meeting point x₁ = x₂, so we can equalize the two equations

½ a₁ t² = x₀₂ + ½ a₂ t²

½ t² (a₁ -a₂) = x₀₂

t =
`\sqrt{ (2 x_(o2) )/( (a_1 - a_2)) }`

let's calculate

t =
`\sqrt{(2 \ 4.04 \ 10^(-2) )/( ( 6.687^(10) + 1.23 10^(14) ) ) }`

t =
`\sqrt{ (8.08 \ 10^(-2) )/( 1.2306 \ 10^(14) ) }`

t = 2.56 10⁻⁸ s

now we can calculate the position

x = ½ a₂ t²

x = ½ 1.23 10¹⁴ (2.56 10⁻⁸)²

x = 4.03 10⁻² m