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

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

To find the magnitude of the electric field, we can use the equation of motion for an object in a constant electric field:

v = u + at

Here, v is the final velocity of the object, u is the initial velocity (which is 0 since the object starts from rest), a is the acceleration (which is given by the electric field), and t is the time interval.

Given:
q = 6.7 µC = 6.7 × 10^-6 C (charge)
m = 9.60 × 10^-5 kg (mass)
v = 2.06 × 10^3 m/s (final velocity)
t = 1.03 s (time interval)

Firstly, we need to calculate the acceleration using the formula:

a = (q/m) * E

Where E is the electric field.

Rearranging the equation, we get:

E = (a * m) / q

Now, we have all the values we need to calculate E.

Substituting the given values:

E = (a * m) / q
E = [(v - u) * m] / (q * t)
E = [(2.06 × 10^3 m/s - 0 m/s) * 9.60 × 10^-5 kg] / [(6.7 × 10^-6 C) * 1.03 s]

Calculating this expression will give us the magnitude of the electric field.