The following problem applies to questions 8 and 9: a glass window acquires a net negative charge on its surface after being cleaned. Particles of dust, which are usually charged positively, start accelerating toward the window. If a particle travels a distance of 1 meter before reaching the window, in a time duration of 10 sec, and if the mass of the particle is 1 micro-gram and the charge on the particle is 10-12 Coulomb, then the magnitude of the electric field intensity is Group of answer choices

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

its c i just checked it