Question

A charged cloud system produces an electric field in the air near the Earth's surface. A particle of charge -2.6 10-9 C is acted on by a downward electrostatic force of 3.0 ✕ 10-6 N when placed in this field.

(a) What is the magnitude of the electric field?

(b) What are the magnitude and direction of the electrostatic force Fel exerted on a proton placed in this field?

(c) What is the magnitude of the gravitational force Fg on the proton?

(d) What is the ratio Fel / Fg in this case?

Answer 1

a) E = 1153.85 N/C

b) Fel = 1.846*10⁻¹⁶N

A proton placed in this field will experience an upward force.

c) Fg = 1.638*10⁻²⁶N

d) Fel / Fg = 1.126*10¹⁰

Explanation:

Given

q = -2.6*10⁻⁹ C

Fel = 3.0*10⁻⁶ N

a) We can use the formula

Fel = q*E ⇒ E = Fel/q

⇒ E = (3.0*10⁻⁶ N)/(2.6*10⁻⁹ C) = 1153.85 N/C

b) We apply the same equation using E = 1153.85 N/C:

Fel = q*E

⇒ Fel = (1.6*10⁻¹⁹C)*(1153.85 N/C) = 1.846*10⁻¹⁶N

A proton placed in this field will experience an upward force.

c) We use the formula

Fg = mp*g

⇒ Fg = (1.67*10⁻²⁷Kg)*(9.81 m/s²) = 1.638*10⁻²⁶N

d) Ratio: Fel / Fg = (1.846*10⁻¹⁶N)/(1.638*10⁻²⁶N) = 1.126*10¹⁰