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Two plates separated by a distance d = 14.8 mm are charged to a potential difference V = 7.25 V . A constant force F = 7.81 N pushes a − 8.30 mC charge from the positively charged plate to the negatively charged plate. Calculate the change in the kinetic energy Δ K , potential energy Δ U , and total energy Δ E of the charge as it travels from one plate to the other. Assume the initial speed of the charge is 0.

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Answer:

  • change in Kinetic Energy Δ K = 55.388 mJ
  • change in potential energy Δ U  = 60.175 mJ
  • total energy ΔE of the charge = 115.563 mJ

Explanation:

Part (a) change in Kinetic Energy Δ K

Δ K = Fd + q(Ed)

but Ed = V and

V is given as 7.25 V

d is given as 14.8 mm

F is given as 7.81 N

q is given as -8.3 mC

Δ K = Fd + q(V)

Δ K = (7.81 * 14.8 X 10⁻³) - 8.3 X 10⁻³ *7.25  

Δ K = 0.055388 J = 55.388 mJ

Part(b) change in potential energy Δ U

ΔU = |q|*V

ΔU = 8.30 X 10⁻³ * 7.25

ΔU = 60.175 mJ

Part(c) total energy ΔE of the charge

ΔE  = Δ K  + ΔU

ΔE  = 55.388 mJ + 60.175 mJ

ΔE  = 115.563 mJ

change in Kinetic Energy Δ K = 55.388 mJ

change in potential energy Δ U  = 60.175 mJ

total energy ΔE of the charge = 115.563 mJ

  • the calculation is as follows:

(a)

Δ K = Fd + q(Ed)

but

Ed = V

and

V is given as 7.25 V

d is given as 14.8 mm

F is given as 7.81 N

q is given as -8.3 mC

Now

Δ K = Fd + q(V)

Δ K[tex]= (7.81 \times 14.8 \times 10^{-3}) - 8.3 \times 10^{-3} \times 7.25[/tex]

Δ K = 0.055388 J

= 55.388 mJ

(B)

[tex]\Delta U = |q|\times V\\\\ = 8.30 \times 10^{-3} \times 7.25[/tex]

ΔU = 60.175 mJ

(c)

ΔE  = Δ K  + ΔU

ΔE  = 55.388 mJ + 60.175 mJ

ΔE  = 115.563 mJ

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