Oppositely charged parallel plates are separated by 5.35 mm. A potential difference of 600 V exists between the plates.

(a) What is the magnitude of the electric field between the plates?

(b) What is the magnitude of the force on an electron between the plates?
N

(c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.94 mm from the positive plate? J

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To find the answers to these questions, we can use some basic principles of electrostatics and potential difference.

(a) The magnitude of the electric field between the plates can be found using the formula:

Electric field = Potential difference / Distance between plates

Given:
Potential difference (V) = 600 V
Distance between plates (d) = 5.35 mm = 0.00535 m

Electric field = 600 V / 0.00535 m ≈ 112,149 V/m

Therefore, the magnitude of the electric field between the plates is approximately 112,149 V/m.

(b) To find the magnitude of the force on an electron between the plates, we can use the formula:

Force = Electric field * Charge

The charge of an electron is -1.6 x 10^-19 Coulombs (C). Since the plates are oppositely charged, the force on an electron will be in the direction of the electric field. Therefore, the force on an electron will be:

Force = (112,149 V/m) * (-1.6 x 10^-19 C)

Note: The minus sign is used to indicate that the force is in the opposite direction to the electric field.

Force ≈ -1.79 x 10^-14 N

Therefore, the magnitude of the force on an electron between the plates is approximately 1.79 x 10^-14 N.

(c) To find the work done on the electron to move it to the negative plate, we can use the formula:

Work = Force * Distance

The force on the electron is the same as calculated in part (b), -1.79 x 10^-14 N. The distance the electron needs to move is the initial distance from the positive plate to the negative plate, minus the initial position of the electron from the positive plate. Therefore:

Distance = (Distance between plates) - (Initial position of the electron)

Distance = 0.00535 m - 0.00294 m ≈ 0.00241 m

Work = (-1.79 x 10^-14 N) * (0.00241 m)

Work ≈ -4.32 x 10^-17 J

Since the electron is moving against the electric field, the work done on the electron will be negative.

Therefore, the amount of work that must be done on the electron to move it to the negative plate is approximately -4.32 x 10^-17 J.

psyDag what a useless pointless answer. thanks for nothing