The electric field strength is 150 times 104 NC inside a par

The electric field strength is 1.50 times 10^4 N/C inside a parallel-plate capacitor with a 1.30 mm spacing. An electron is released from rest at the negative plate. What is the electron\'s speed when it reaches the psitive plate? Express your answer with the appropriate units.

Solution

F = q E

m a = q E

a = q E / m

a = (1.6 * 10^-19 * 1.50 * 10⁴ ) / ( 9.11 * 10^-31)

a = 2.63 * 10¹⁵ m/s²

v² = u² + 2 a s

v² = 0 + ( 2 * 2.63 * 10¹⁵ * 1.30 * 10^-3 )

v = 2.61 * 10⁶ m/s