Question

g A small object of mass 2.5 g and charge 18 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What is the magnitude and direction of the electric field

Answer 1

Answer:
`1361.11\ N/C,\text{upward}`

Step-by-step explanation:

Given

Mass of particle is
`m=2.5\ gm`

Charge of particle is
`q=18\ \mu C`

Electrostatic force must balance the weight of the particle

`\lim_(n \to \infty) a_n \Rightarrow mg=qE\\\\\Rightarrow E=(2.5* 9.8* 10^(-3))/(18* 10^(-6))\\\\\Rightarrow E=1361.1\ N/C`

Direction of the electric field is in upward direction such that it opposes the gravity force.