a.)A pair of oppositely charged parallel plates are separated by 5.58 mm. A pote

Question

a.)A pair of oppositely charged parallel plates are separated by 5.58 mm. A potential difference of 585 V exists between the plates. What is the strength of the electric field
between the plates?
Answer in units of V/m

019 (part 2 of 3) 10.0 points
What is the magnitude of the force on an
electron between the plates?
Answer in units of N

020 (part 3 of 3) 10.0 points
How much work must be done on the electron
to move it to the negative plate if it is initially
positioned 2.97 mm from the positive plate?
Answer in units of J

Explanation / Answer

For the given set of plates, V = 585 V and d = 5.58 × 103 m. Therefore,
the magnitude of the electric field between the plates is

E= V/d = 585/5.58 × 103 = 1.05 V/m = 1.05 N/C

(b) The magnitude of the force on an electron located between the plates is
F = q * E = eE = (1.60 × 1019 C)(1.05 × 105 N/C) = 1.68 x 1014 N

(c) The displacement of the electron as it moves from the initial position to the
negative plate is
s = d 2.97 mm= 5.58 mm 2.97 mm = 2.61 mm

An applied force with magnitude F = eE = 1.68 × 1014 N, directed toward the
negative plate, must be used to offset the influence of the field and move the
electron without acceleration. The work done by the applied force as the
electron is moved to the negative plate is

W = F*s*cos0° = (1.68 × 1014 N)(2.61 × 103 m)(1.00) = 4.38 × 1017 J

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