# Velocity of an electron within a weak electric field?

• islimst3ri
In summary: When you are using vectors, the direction of the vector is always positive, regardless of the sign of the magnitude. So in this case, the E-field is positive and the g-field is negative.

#### islimst3ri

This is the question:

"An electron is released from rest in a weak electric field given by vector E = -1.80 10-10 N/C jhat. After the electron has traveled a vertical distance of 1.2 µm, what is its speed? (Do not neglect the gravitational force on the electron.)"

So far, the only approach I can think of is by using the equation Δy= 1/2 a*t. Since a = (e*E)/m and t = Δx / v(not), I have:

Δy = 1.2 µm = (.5) * [ (1.6E-19 C * -1.8E-10 N/C) / 9.11E-31 kg ] * (Δx / v)^2.

Since I have two unknowns (the Δx and v), I have been stuck for a while now. If anyone could offer any insight it would be greatly appreciated. This problem has been driving me mad! Thanks in advance for any efforts or hints!

You can't use (delta x)/v=(delta t) unless the motion has uniform velocity. Not true here. You can either use F=ma to analyze the motion or you can use conservation of energy to do it more easily.

Dick,

Maybe I'm missing something here but I don't see it, what do you mean by using the conservation of energy? Are you suggesting that I should use the kinetic form, by setting mgh = 1/2 mv^2? If so I don't understand why the Electric field charge could be involved at all. Please explain a little bit further as I am not the brightest of physics students. Thank you.

welcome to pf!

hi islimst3ri! welcome to pf!

(try using the X2 icon just above the Reply box )
islimst3ri said:
… what do you mean by using the conservation of energy? Are you suggesting that I should use the kinetic form, by setting mgh = 1/2 mv^2? If so I don't understand why the Electric field charge could be involved at all.

all fields supply energy

the energy is the work done by the field: force "dot" displacement

for gravity, it's mgh (force mg, displacement h)

for electric fields, it's qEh (force qE, displacement h)

so use the work energy theorem … change in KE = work done by the fields

tiny-tim said:
so use the work energy theorem … change in KE = work done by the fields

When I follow this approach, I have get
W = KE = qEh
= (1.6E-19 C)(-1.8E-10 N/C)(1.2E-6 m)
= -3.46E-35 J

After finding work, I plugged it into W = 1/2 mv2 as follows:

-3.46E-35 J = (.5)(9.1E-31 kg)(v2)
-6.91E-35 J = (9.1E-31 kg)(v2)
-7.60E-5 J/kg = (v2)

Since you cannot take a square-root of a negative number, I am once again stuck staring at this problem. Is there something that I'm missing or did wrong on the steps above? Thank you guys for being patient with me.

what happened to gravity?

(and don't forget electrons are negative)

Tim,

So WebAssign is down right now thus I can't check my answer. But I'll show my work and could you please scan over it to see if its right. A simple yes or no will suffice.

WE-Field = KE = qEh
= -(1.6E-19 C)(-1.8E-10 N/C)(1.2E-6 m)
= 3.46E-35 J

Wgravity = mgh
= (9.1E-31 kg)(9.81 m/s2)(1.2E-6 m)
= 1.07E-35 J

Using W = 1/2 mv2 for both E-field and gravitational field, I got v = .00872 and .00485 m/s, respectively. I then subtracted one from the other and get .00386 m/s, or 3.86 mm/s (the units WebAssign was looking for). How does this look?

Is the electron rising or falling in the electric field? The gravitational field?

Since the electric field is moving with the vector in the direction of j-hat, it is moving upwards, while the gravitational field, intuitively, is moving downward. Since the electron was able to travel vertically for 1.2 µm, this indicates that the E-field is stronger than the g-field, thus this was why I subtracted the g-field speed from the e-field speed to get the electron's speed. However, I submitted this and still got the wrong answer. What am I doing wrong?

islimst3ri said:
Since the electric field is moving with the vector in the direction of j-hat, it is moving upwards, while the gravitational field, intuitively, is moving downward. Since the electron was able to travel vertically for 1.2 µm, this indicates that the E-field is stronger than the g-field, thus this was why I subtracted the g-field speed from the e-field speed to get the electron's speed. However, I submitted this and still got the wrong answer. What am I doing wrong?

Start by making clear what your coordinate system is. Assign directions to $$(\hat{i},\hat{j},\hat{k})$$. What does the negative sign mean for the E field magnitude? What sign should you assign to g if it's a vector?

u r making mistakes in the signs
otherwise the method looks quite fine to me...

## 1. What is the velocity of an electron within a weak electric field?

The velocity of an electron within a weak electric field depends on the strength of the electric field and the mass of the electron. In general, the velocity will increase as the electric field strength increases.

## 2. How is the velocity of an electron affected by a weak electric field?

In a weak electric field, the velocity of an electron will be relatively low compared to strong electric fields. This is because the force exerted on the electron is not strong enough to significantly accelerate it.

## 3. What is the relationship between the velocity of an electron and its kinetic energy in a weak electric field?

The velocity and kinetic energy of an electron are directly proportional in a weak electric field. This means that as the velocity increases, so does the kinetic energy.

## 4. Can the velocity of an electron in a weak electric field be negative?

Yes, the velocity of an electron in a weak electric field can be negative. This indicates that the electron is moving in the opposite direction of the electric field.

## 5. How does the velocity of an electron within a weak electric field compare to its speed of light?

The velocity of an electron in a weak electric field is much lower than the speed of light. In fact, the speed of light is the maximum possible velocity for any object, including electrons, in the universe.