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An electron has an initial velocity of

1 × 106 m/s in the x direction. It enters a
uniform electric field E~ = (360 N/C) ˆj which
is in the y direction.
Find the acceleration of the electron. The
fundamental charge is 1.602 × 10−19 C and
the mass of the electron is 9.109 × 10−31 kg.
Answer in units of m/s
2
. Your answer must
be within ± 5.0%

User Macou
by
8.3k points

1 Answer

3 votes

To find the acceleration of the electron in the given electric field, we can use the equation of motion for a charged particle in an electric field.

Given:

Initial velocity of the electron, v_initial = 1 × 10^6 m/s (in the x-direction)

Electric field, E~ = (360 N/C) ˆj (in the y-direction)

Fundamental charge, e = 1.602 × 10^(-19) C

Mass of the electron, m = 9.109 × 10^(-31) kg

The force experienced by a charged particle in an electric field is given by the equation:

F~ = qE~

Where:

F~ is the force vector,

q is the charge, and

E~ is the electric field vector.

Since the electron has a negative charge, the force vector F~ will be in the opposite direction of the electric field vector E~.

The force acting on the electron can be calculated as:

F~ = -qE~

Substituting the values:

F~ = -(1.602 × 10^(-19) C) × (360 N/C) ˆj

Now, we can use Newton's second law of motion, F~ = ma~, to find the acceleration of the electron.

Since the force acting on the electron is in the y-direction (opposite to the electric field direction), and the mass of the electron is known, we have:

F_y = ma_y

Substituting the force in the y-direction:

-(1.602 × 10^(-19) C) × (360 N/C) = (9.109 × 10^(-31) kg) × a_y

Solving for a_y:

a_y = -[(1.602 × 10^(-19) C) × (360 N/C)] / (9.109 × 10^(-31) kg)

Calculating the value:

a_y ≈ -6.56 × 10^11 m/s^2

The acceleration of the electron is approximately -6.56 × 10^11 m/s^2 in the y-direction. Note that the negative sign indicates that the acceleration is opposite to the direction of the electric field, as expected for a negatively charged particle.

User Kayak
by
8.0k points