28: Problem 28.2

INTRO:
The electric field strength is 2.50×10⁴ N/C inside a parallel-plate capacitor with a 1.50 mm spacing. An electron is released from rest at the negative plate.

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PART A:
What is the electron's speed when it reaches the positive plate?
Express your answer with the appropriate units.

SOLUTION:
Givens / conversions:
E = 2.50×10⁴ N/C 
d =1.50 mm = 0.0015 m

we also need to know the properties of an electron
q_e = -1.60×10^-19 C
m_e = 9.11×10^-31 kg

Using the energy equations:
ΔPE = U_f - U_i = qEd₂ - qEd₁
since it goes from x = d to x = 0,

ΔPE = qEd​

ΔKE = K_f - K_i = ½m(v_f² - v_i²)
since we know that it starts at rest... 

ΔKE = ½mv_f²​

ΣEnergy = 0: ΔPE + ΔKE = 0
qEd + ½mv_f² = 0

now solve for v_f... 
v_f = [-2qEd/m]^½

plug in values... 
v_f = [-2(-1.60×10^-19 C)⋅(2.50×10⁴ N/C )⋅(0.0015 m)/(9.11×10^-31kg)]^½

⇒ ∴ v_f =​ 3.629×10⁶ m/s

v_f =​ 3.63×10⁶ m/s
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