Answer:
the magnitude of the electric field intensity is 20 N/C
Explanation:
Given the data in the question;
mass m = 1 micro gram = 1 × 10⁻⁹ kg
time duration t = 10 sec
distance s = 1 m
the charge on the particle q = 10⁻¹² Coulomb
force applied on a charged particle due to electric field E is;
F = Eq ------ equ 1
where q is the charge on the particle.
Also, force on a particle with mass m will be;
F = ma ------ equ
where a is acceleration
so F = ma = Eq
ma = Eq -------- equ 3
using kinetic equation
Distance = 1/2×at²
where a is acceleration and t is the time period
now lets consider that initial velocity is zero (0)
Here;
1 m = 1/2 × a × ( 10 s )²
1 m = a × 50 s²
a = 1 m / 50 s²
a = 0.02 m/s²
so, from equation 3
ma = Eq
E = ma / q
we substitute
E = (1 × 10⁻⁹ kg × 0.02) / 10⁻¹² Coulomb
E = 2 × 10⁻¹¹ / 10⁻¹²
E = 20 N/C
Therefore, the magnitude of the electric field intensity is 20 N/C
