The acceleration rate is
a = 2.06*10^3/1.03 = 2000 m/s^2
The force on the particle is
M*a = 0.192 Newtons
That equals Q*E, where E is the field strength and Q is the charge in Coulombs.
E = (0.192 N)/6.7*10^-6C
= 2.86*10^4 N/C
A small object, which has a charge q = 6.7 µC and mass m = 9.60 10-5 kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.06 103 m/s in a time of 1.03 s. Determine the magnitude of the electric field.
Answer in N/C
1 answer