Motion of a charged particle: A pair of charged conducting plates produces a uniform field of 12,000 N/C, directed to the right, between the plates. The separation of the plates is 40 mm. An electron is projected from plate A, directly toward plate B, with an initial velocity of v_o = 1.0 × 10⁷m/s, as shown in the figure. (e= 1.60 × 10^-19C, ε₀ = 8.85 × 10^-12 C²/N ∙ m², m_el = 9.11 × 10^-31 kg) The distance of closest approach of the electron to plate B is nearest to A) 16 mm. B) 18 mm. C) 20 mm. D) 22 mm. E) 24 mm.

Question

Quizplus · Accepted Answer

We can use the equations of motion for a charged particle in an electric field.

The force on the electron, F = qE, where q is the charge of the electron and E is the electric field.

The acceleration of the electron, a = F/m, where m is the mass of the electron.

Using the initial velocity and acceleration in the horizontal direction, we get the equation of motion:

x = vt + (1/2)at^2

where x is the distance of closest approach, v is the initial velocity, t is the time taken to reach the distance of closest approach, and a is the acceleration.

Since the electron is projected directly towards plate B, the time taken to reach the distance of closest approach is the same as the time taken to travel half the distance between the plates.

Therefore, we can calculate the time taken using the equation:

t = (d/2)/v

where d is the distance between the plates.

Plugging in the values, we get:

t = (0.04 m)/(1.0 × 10^7 m/s) = 4 × 10^-9 s

Now, we can use the equation of motion to calculate the distance of closest approach:

x = vt + (1/2)at^2

where v is the initial velocity, a is the acceleration, and t is the time taken.

The initial velocity is directed towards plate B, and the acceleration is in the opposite direction.

Therefore, we have:

a = -qE/m = -(1.60 × 10^-19 C)(1.2 × 10^4 N/C)/(9.11 × 10^-31 kg) = -2.11 × 10^14 m/s^2

Plugging in the values, we get:

x = (1.0 × 10^7 m/s)(4 × 10^-9 s) + (1/2)(-2.11 × 10^14 m/s^2)(4 × 10^-9 s)^2 = 16.8 mm

Therefore, the distance of closest approach is nearest to 16 mm, which corresponds to choice A.

Motion of a Charged Particle: a Pair of Charged Conducting