Question

A small object of mass 3.66g and charge-19 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What ares the magnitude and direction of the electric field? magnitude N/C direction

Answer 1

Step-by-step explanation:

It is given that,

Mass of the object, m = 3.66 kg

Charge, q = -19 μC = -19 × 10⁻⁶ C

It is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground such that,

`F_g=F_e`

`F_g\ and\ F_e` are gravitational and electrostatic forces respectively

`mg=qE`

`E=(mg)/(q)`

`E=(3.66\ kg* 9.8\ m/s^2)/(-19* 10^(-6)\ C)`

E = −1887789.47 N/C

`E=-1.89* 10^6\ N/C`

Negative sign shows that the electric field is in the opposite direction of the electric force. Since, the weight of the object is in downward direction and its electric force (which is balancing its weight) is in upward direction. So, we can say that the electric field is in downward direction.