Answer:
3.53 N/C
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
Answer: The magnitude of the electric field is 3.53 N/C
Explanation: Please see the attachments below
