In this example we will calculate the work done on an electron moving in a unifo
ID: 1574744 • Letter: I
Question
In this example we will calculate the work done on an electron moving in a uniform electric field. Two large conducting plates separated by 6.36 mm carry charges of equal magnitude and opposite sign, creating a constant electric field with magnitude 2.80×103N/C between the plates. An electron moves from the negatively charged plate to the positively charged plate. How much work does the electric field do on the electron?
The above image shows our sketch. The electric field is directed from the positive plate toward the negative plate. F E=qE , so for an electron with negative charge q=e, the electric force F E points in the direction opposite to the electric field. Its magnitude is FE=eE. The electric field is uniform, so the force it exerts on the electron is constant during the electron’s motion.
The force and displacement are parallel; the work W done by the electric-field force during a displacement of magnitude d is W=Fedcos with =0, so
W = Fed = eEd
= (1.60×1019C)(2.80×103N/C)(6.36×103m)
= 2.85×1018J
The amount of work done is proportional to the electric-field magnitude E and to the displacement magnitude d of the electron. The electric field does positive work on the electron. If there are no other forces, the electron's kinetic energy increases by the same amount as the work done on the electron by the electric field.
In the same problem, how much work does the electric field do on the electron if the magnitude of the field is increased to 6.00×103N/C, and the separation between the plates is reduced to 3.30 mm?
E-280 × 1 N/C d 636 mm -eExplanation / Answer
W = Fe*d
Fe = q*E
W = q*E*d
Using given values:
W = 1.6*10^-19*6*10^3*3.3*10^-3
W = 3.17*10^-18 J
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