A uniform electric field of magnitude 395 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.40 cm. What is the work done by the field on the electron? What is the change in potential energy associated with the electron? What is the velocity of the electron?
Work=Eqd
velocity= work=1/2 mv^2
1/2 mv^2=Eqd solve for v
To find the work done by the electric field on the electron, we can use the formula:
Work = Force x Displacement x cos(theta)
Where:
- Force is the magnitude of the electric field acting on the electron, given as 395 N/C.
- Displacement is the distance the electron has moved, given as 3.40 cm.
- theta is the angle between the direction of the force and the displacement. Since the electric field and displacement are both in the positive x-direction, theta is 0 degrees.
Let's calculate the work done by the electric field:
Work = (395 N/C) x (3.40 cm) x cos(0°)
First, we need to convert the displacement from centimeters to meters:
Displacement = 3.40 cm = 3.40 x 10^-2 m
Now we can calculate the work:
Work = (395 N/C) x (3.40 x 10^-2 m) x cos(0°)
= (395 N/C) x (3.40 x 10^-2 m) x 1
= 13.43 J
Therefore, the work done by the electric field on the electron is 13.43 J.
Next, let's find the change in potential energy associated with the electron. The change in potential energy is equal to the work done by the electric field:
Change in Potential Energy = Work
= 13.43 J
Therefore, the change in potential energy associated with the electron is 13.43 J.
Now, to determine the velocity of the electron, we can use the conservation of energy principle. The work done by the field is equal to the change in kinetic energy of the electron.
Work = Change in Kinetic Energy
13.43 J = (1/2)mv^2
Where:
- m is the mass of the electron, given as 9.11 × 10^-31 kg.
- v is the velocity of the electron (what we want to find).
Solving for v:
v^2 = (2 * 13.43 J) / (9.11 × 10^-31 kg)
v^2 = 29.39 x 10^30 m^2/s^2
Taking the square root of both sides:
v = sqrt(29.39 x 10^30 m^2/s^2)
v = 5.42 x 10^15 m/s
Therefore, the velocity of the electron is approximately 5.42 x 10^15 m/s.