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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

Respuesta :

nuhulawal20

Answer:

the magnitude of the electric field intensity is 20 N/C

Explanation:

Ā Given the data in the question;

mass m = 1 micro gram = 1 Ɨ 10ā»ā¹ kg

time duration t = 10 sec

distance s = 1 m

the charge on the particle q = 10ā»Ā¹Ā² Coulomb

force applied on a charged particle due to electric field E is;

F = Eq Ā ------ equ 1

where q is the charge on the particle.

Also, force on a particle with mass m Ā will be;

F = ma Ā ------ equ

where a is acceleration

so F = ma = Eq

ma = Eq -------- equ 3

using kinetic equation

Distance = 1/2ƗatĀ²

where a is acceleration and t is the time period

now lets consider that initial velocity is zero (0)

Here;

1 m = 1/2 Ɨ a Ɨ ( 10 s )Ā²

1 m = a Ɨ 50 sĀ²

a = 1 m / 50 sĀ²

a = 0.02 m/sĀ²

so, from equation 3

ma = Eq

E = ma / q

we substitute

E = (1 Ɨ 10ā»ā¹ kg Ɨ 0.02) / 10ā»Ā¹Ā² Coulomb

E = 2 Ɨ 10ā»Ā¹Ā¹ / 10ā»Ā¹Ā²

E = 20 N/C

Therefore, the magnitude of the electric field intensity is 20 N/C

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