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 = 2.0 × 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 velocity of the electron as it strikes plate B is closest to

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the equations of motion and also consider the force acting on the electron due to the electric field between the plates.

First, we can use the equation of motion:

v = u + at

where v is the final velocity (which we want to find), u is the initial velocity (given as 2.0 × 10S1U1P17S1S1P0 m/s), a is the acceleration, and t is the time it takes for the electron to travel between the plates.

We can find the acceleration using the force equation:

F = ma = qE

where F is the force on the electron, m is its mass, q is its charge (given as -1.60 × 10S1U1P1-19S1S1P0 C), and E is the electric field strength (given as 12,000 N/C).

Rearranging this equation, we get:

a = qE/m

Plugging in the numbers, we get:

a = (-1.60 × 10S1U1P1-19S1S1P0 C)(12,000 N/C)/(9.11 × 10S1U1P1-31S1S1P0 kg)

a = -2.18 × 10S1U1P13S1S1P0 m/sS1U1P2S1S1P0

Note that the negative sign indicates that the force and acceleration are in the opposite direction to the initial velocity of the electron.

Now we can find the time it takes, t, for the electron to travel between the plates. We know that the distance between the plates is 40 mm or 0.04 m. Also, since the electric field is constant, the acceleration is also constant, leading to uniform acceleration motion, allowing us to determine the time as follows:

s = ut + 1/2 atS1U1P2S1S1P0

0.04 m = (2.0 × 10S1U1P17S1S1P0 m/s)t + 1/2 (-2.18 × 10S1U1P13S1S1P0 m/sS1U1P2S1S1P0)tS1U1P2S1S1P0

Solving, we get:

t = 3.74 × 10S1U1P-12S1S1P0 s

Now we can use the equation of motion again to find the final velocity:

v = u + at

v = (2.0 × 10S1U1P17S1S1P0 m/s) + (-2.18 × 10S1U1P13S1S1P0 m/sS1U1P2S1S1P0)(3.74 × 10S1U1P-12S1S1P0 s)

v = 1.47 × 10S1U1P17S1S1P0 m/s

So the closest answer choice is (B) 1.5 × 10S1U1P17S1S1P0 m/s.

A Pair of Charged Conducting Plates Produces a Uniform Field