Question

What are the magnitude and direction of a uniform electric field perpendicular to the ground that is able to suspend a particle of mass m = 1.30 g carrying a charge of +6.90 µC in midair, assuming gravity and the electrostatic force are the only forces exerted on the particle?

Answer 1

the magnitude and direction of the uniform electric field is 1846.38 N/C and UPWARD respectively

Step-by-step explanation:

Given the data in the question;

mass m = 1.30 g = 1.30 × 10⁻³ kg

Charge q = +6.90 µC = 6.90 × 10⁻⁶ C

we know that; g = 9.8 m/s

assuming gravity and the electrostatic force are the only forces exerted on the particle, hence the relation is;

F = mg and F = Eq

so

mg = Eq

make E subject of formula

E = mg / q

so we substitute

E = [ (1.30 × 10⁻³) × 9.8 ] / 6.90 × 10⁻⁶

E = 0.01274 / 6.90 × 10⁻⁶

E = 1846.38 N/C

Since the charge is positive ( + ), { direction is Upward }

Therefore, the magnitude and direction of the uniform electric field is 1846.38 N/C and UPWARD respectively