An electron is accelerated by a uniform electric field (1000 V/m) pointing vertically upward. Use energy methods to get the magnitude and direction of its velocity after it moves 0.10 cm form rest. Does the electron gain or lose potential energy?

To determine the magnitude and direction of the electron's velocity after it moves and whether it gains or loses potential energy, we can use energy principles.

First, we need to calculate the change in potential energy undergone by the electron. The potential energy of a charged particle in a uniform electric field is given by the equation:

ΔPE = q * ΔV,

where ΔPE is the change in potential energy, q is the charge of the particle, and ΔV is the change in voltage.

In this case, the charge of the electron (q) is equal to the elementary charge, e, which is approximately 1.6 x 10^(-19) C. The change in voltage (ΔV) is given as 1000 V/m, and the change in position (Δx) is equal to 0.10 cm, which we need to convert to meters:

Δx = 0.10 cm = 0.10 x 10^(-2) m = 1 x 10^(-3) m.

Now, let's calculate the change in potential energy:

ΔPE = (1.6 x 10^(-19) C) * (1000 V/m) * (1 x 10^(-3) m)
= 1.6 x 10^(-22) Joules.

Since the electric field is pointing vertically upward, and the electron moves from rest to a higher potential position, we know that the electron gains potential energy.

Now, using the principle of conservation of energy, we can equate the change in potential energy to the change in kinetic energy:

ΔKE = ΔPE.

The kinetic energy of an object is given by the equation:

KE = (1/2) * m * v^2,

where KE is the kinetic energy, m is the mass of the object, and v is its velocity.

The mass of an electron (m) is approximately 9.1 x 10^(-31) kg.

Since the electron was initially at rest, the initial kinetic energy is zero:

KE_initial = 0.

Hence, the change in kinetic energy (ΔKE) is equal to the final kinetic energy:

ΔKE = KE_final.

Setting up the equation:

(1/2) * m * v^2 = 1.6 x 10^(-22) Joules.

Substituting the values:

(1/2) * (9.1 x 10^(-31) kg) * v^2 = 1.6 x 10^(-22) Joules.

Solving for v:

v^2 = (2 * (1.6 x 10^(-22) Joules)) / (9.1 x 10^(-31) kg)
= 3.52 x 10^8 m^2/s^2.

Taking the square root of both sides:

v ≈ 5.93 x 10^4 m/s.

Therefore, the magnitude of the electron's velocity after moving 0.10 cm is approximately 5.93 x 10^4 m/s upward.

In summary, the electron gains potential energy, and its velocity is approximately 5.93 x 10^4 m/s upward.

To determine the magnitude and direction of the electron's velocity after moving 0.10 cm, we can use energy methods.

1. First, let's calculate the potential energy gained or lost by the electron. The potential energy change is given by the formula:

∆PE = q∆V

where q is the charge of the electron and ∆V is the change in electric potential.

For an electron, the charge (q) is -1.6 x 10^-19 C (Coulombs).

∆V is the change in electric potential, which is the product of the electric field strength (E) and the distance (d) the electron moves:

∆V = E * d

Substituting the given values:
E = 1000 V/m,
d = 0.10 cm = 0.001 m,

∆V = (1000 V/m) * (0.001 m) = 1 V

∆PE = (-1.6 x 10^-19 C) * (1 V) = -1.6 x 10^-19 J

Therefore, the electron loses 1.6 x 10^-19 J of potential energy.

2. Now, let's use the principle of conservation of energy to find the kinetic energy gained by the electron. Since there is no work done against any other forces (e.g., friction), the potential energy lost is converted into kinetic energy gained. Therefore, the magnitude of the kinetic energy will be equal to the magnitude of the potential energy loss:

KE = |∆PE| = 1.6 x 10^-19 J

3. Lastly, we can use the kinetic energy formula to find the magnitude of the electron's velocity:

KE = 0.5 * m * v^2

where m is the mass of the electron and v is its velocity.

The mass (m) of an electron is approximately 9.11 x 10^-31 kg.

Rearranging the formula, we get:

v^2 = (2 * KE) / m

v^2 = (2 * 1.6 x 10^-19 J) / (9.11 x 10^-31 kg)

v^2 ≈ 3.50 x 10^11 m^2/s^2

v ≈ √(3.50 x 10^11 m^2/s^2)

v ≈ 5.91 x 10^5 m/s

Therefore, the magnitude of the electron's velocity after moving 0.10 cm is approximately 5.91 x 10^5 m/s.

Since the electric field is pointing vertically upward, the electron gains potential energy as it moves against the electric field.

E=Δφ/Δx

PE->W(work of electric field) -> KE
W=e•Δφ=e•E•Δx
KE= mv²/2
mv²/2= e•E•Δx
v=sqrt{2•e•E•Δx/m}=
=sqrt{2•1.6•10⁻¹⁹•1000•0.001/9.1•10⁻³¹}=
=5.93•10⁵ m/s.