A pair of oppositely charged parallel plates are separated by 5.58 mm. A potenti

Question

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.

016 (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.

017 (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

Work to move charge between plates   W = V q = E qd   (force on charge X distance moved) E = V / d = 585 / .00558 = 1.008 * 10E5 N / C F = E q = 1.0088 * 10E5 * 1.6 * 10E-19 = 1.61 * 10E-14 N W = F d = 1.61 * 10E-14 * (0.00558 -0.00297) = 4.20 * 10E-17J Work to move charge between plates   W = V q = E qd   (force on charge X distance moved) E = V / d = 585 / .00558 = 1.008 * 10E5 N / C F = E q = 1.0088 * 10E5 * 1.6 * 10E-19 = 1.61 * 10E-14 N W = F d = 1.61 * 10E-14 * (0.00558 -0.00297) = 4.20 * 10E-17J Work to move charge between plates   W = V q = E qd   (force on charge X distance moved) E = V / d = 585 / .00558 = 1.008 * 10E5 N / C F = E q = 1.0088 * 10E5 * 1.6 * 10E-19 = 1.61 * 10E-14 N W = F d = 1.61 * 10E-14 * (0.00558 -0.00297) = 4.20 * 10E-17J

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