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 = 1.20 g carrying a charge of +4.00 µC in midair, assuming gravity and the electrostatic force are the only forces exerted on the particle? answer in N/C

Answer 1

$\large Fe = weight = mg$
$\large F_e=1.2\times 10^{-3}kg\times 9.81m/s^2 = 1.177 \times 10^{-2} N (upwards)$
Electric Field strength

$\large E= \frac{F_e}{Q}$

$\large E= \frac{ 1.177 \times 10^{-2} N }{4\times 10^{-6}C}$

$\large E= 2.943\times 10^3N/C$

Magnitude of electric field $\large E= 2.943\times 10^3N/C$

Upward