Question
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ = 1.68 10-7 C/m2, and the plates are separated by a distance of 1.42 10-2 m. How fast is the electron moving just before it reaches the positive plate?
Answer
Electric field between the plates of the capacitor = σ/εo N/C
Force acting on charge q = qσ/εo N
Work done on moving this charge across distance Δx = q(σ/εo)Δx
This work gets converted into kinetic energy of the charge having mass m
=> (1/2) mv^2 = q(σ/εo)Δx
=> v = √[(2q(σ/εo)Δx)/m] ... [As stated by you]
Now, plugging values,
mass of electron, m = 9.10938188 × 10-31 kg and
charge of electron, q = 1.60217646 × 10-19 C
=> velocity, v
= √[2 * (1.60217646 × 10-19) * (1.68 x 10^-7 / 8.854 x 10-12) * (1.42 x 10^-2 / 9.10938188 × 10-31] m/s
= √[0.94778 x 10^14)
= 9.74 x 10^6 m/s.
Electric field between the plates of the capacitor = σ/εo N/C
Force acting on charge q = qσ/εo N
Work done on moving this charge across distance Δx = q(σ/εo)Δx
This work gets converted into kinetic energy of the charge having mass m
=> (1/2) mv^2 = q(σ/εo)Δx
=> v = √[(2q(σ/εo)Δx)/m] ... [As stated by you]
Now, plugging values,
mass of electron, m = 9.10938188 × 10-31 kg and
charge of electron, q = 1.60217646 × 10-19 C
=> velocity, v
= √[2 * (1.60217646 × 10-19) * (1.68 x 10^-7 / 8.854 x 10-12) * (1.42 x 10^-2 / 9.10938188 × 10-31] m/s
= √[0.94778 x 10^14)
= 9.74 x 10^6 m/s.
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