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?

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

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

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2