Answer:
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
