Question

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Answer 1

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Answer 2

If the electron misses the upper plate, the magnitude of the electric field is equal to 171.88 Newton per coulomb.

Given the following data:

• Distance = 2 cm to m = 0.02 meter.

• Vertical speed = $1.6 \times 10^6$ m/s

• Vertical distance = 1 cm =  $\frac{0.01}{2} = 0.005\;m$

Scientific data:

• Mass of electron = $9.1 \times 10^{-31}\;kg$

• Charge of electron = $1.6 \times 10^{-19}\;C$

To calculate the magnitude of the electric field:

First of all, we would determine the time taken by this electron to travel through the plates.

Time in the vertical direction.

Mathematically, time is given by this formula:

$Time = \frac{distance}{speed} \\ \\ Time = \frac{0.02}{1.10 \times 10^6} \\ \\ Time = 1.82 \times 10^{-8}\;m/s$

Next, we would find the acceleration of the electron in the vertical direction by using this formula:

$a=\frac{2y}{t^2} \\ \\ a=\frac{2 \times 0.005}{(1.82 \times 10^{-8})^2}\\ \\ a=\frac{0.01}{3.31 \times 10^{-16}}\\ \\ a=3.02 \times 10^{13}\;m/s^2$

The formula for electric field.

Mathematically,  the electric field is given by this formula:

$E=\frac{ma}{q}$

Where:

• q is the charge.

• a is the acceleration.

• m is the mass.

Substituting the given parameters into the formula, we have;

$E=\frac{9.1 \times 10^{-31} \times 3.02 \times 10^{13}}{1.6 \times 10^{-19}}\\ \\ E=\frac{2.75 \times 10^{-17}}{1.6 \times 10^{-19}}$

Electric field, E = 171.88 N/C.

Read more on electric field here: https://brainly.com/question/14372859

Complete Question:

An electron is projected with an initial speed v0 = $1.6 \times 10^6$ m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C

(a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.