Question

A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.8 cm away, in a time 3.5 10-7 s. What is the magnitude of the electric field

Answer 1

Answer: The magnitude of the electric field is 3.53 N/C

Explanation: Please see the attachments below

Answer 2

3.53 N/C

Step-by-step explanation:

Electric field = F / q where F is the force in N and q is charge on the electron

F = mass of an electron × a ( acceleration in m/s)

using equation of motion to solve for the acceleration

s ( distance ) = ut + 0.5 at² since the electron is starting from rest then ut = 0

2s / t² = a

F = me × ( 2s / t²)

E electric field = me × ( 2s / t²) / q = me × 2s / ( t² × q)

me, mass of an electron = 9.11 × 10⁻³¹ kg

E = (9.11 × 10⁻³¹ kg × 2 × 0.038 m) / ( (3.5 × 10⁻⁷s)² × 1.6 × 10⁻¹⁹ C) = 0.0353 × 10² N/C = 3.53 N/C