Ask question

Ask question

All categories

2 years ago

5

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

tarting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 102 N/C and separation between the charged plates is 2.0 cm. Determine the horizontal distance travelled by the electron when it hits the plate.

1 answer:

max2010maxim [7]2 years ago

4 0

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

You might be interested in

Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed

olga55 [171]

Initially, the energies are:

$U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\
=K_{i}=\frac{1}{2} m v_{0}^{2}$

At final point, the energies are:

$U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\
K_{f}=\frac{1}{2} m(0)^{2}=0$

Using conservation law of energy,

$-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\
-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}$

The equation is further simplified as:

$r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\
h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\
&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\
& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}$

7 0

1 year ago

Which statements describe a situation with a displacement of zero? Check all that apply. traveling south for 30 miles, then turn

velikii [3]

<span>The term "displacement" includes a change of position or change in an innate characteristic. The first option would have someone travel in an L-shape, which definitely is a change in position from the starting point. The second option of Ferris wheel with the same entrance and exit does not involve overall displacement since a person would return to the same place they began. The third option of walking around the block does not involve overall displacement since, again, the person would return to the same place they began. The fourth option of an escalator ride does involve overall displacement because a person would finish their journey in a different vertical location from where they started. The last option does not involve overall displacement because one lap around a track will return you to the same place you began.</span>

5 0

2 years ago

Read 2 more answers

It is easier to climb up a slanted slope than a vertical slope

V125BC [204]

IT IS EASIER TO CLIMB A SLANTED SLOPE

3 0

1 year ago

Read 2 more answers

The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

s344n2d4d5 [400]

Answer:

v = √2G $M_{earth}$ / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

Eo = K + U = ½ m1 v² - G m1 m2 / r1

Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

Ef = = - G m1 m2 / R

The mechanical energy is conserved

Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

R = int ⇒ Em = 0

6 0

1 year ago

Assuming motion is on a straight path, what would result in two positive components of a vector?

Aliun [14]

Answer: option <span>A) a train moving north of east at an angle of 25°
</span>

Explanation:

1) You need to choose your axis. In this case North is vertical and positive, South is vertical and negative, East is horizontal and positive, and West is horizontal and negative.

2) The vector with the two positive components is a vector in the first quadrant (North and East). That is what North of East 25° means.

3) Regarding the other options:

<span>B) a bus moving North of East at an angle of 95°: since the angle is greater tnan 90° the vector is in the second quadrant: its horizontal component is negative.
</span>
<span /><span>
</span><span> C) a boat moving South of West at an angle of 40°: this is in the third quadrant: the two components are negative.</span>
<span /><span>
</span><span> D) a car moving south of west at an angle of 10°: as in the option C), this is in the third quadrant: the two components are negative.
</span>

3 0

1 year ago

Read 2 more answers