Question

The electric field required to keep a water drop of mass m just to remain suspended when charged with one electron:
A. mg
B. $\dfrac{em}{g}$
C. emg
D. $\dfrac{mg}{e}$

Answer 1

Hint:A physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is referred to as an electric field. It can also refer to a system of charged particles' physical field. Electric fields are created by electric charges or magnetic fields that change over time. The electromagnetic force, one of nature's four basic forces (or interactions), manifests itself in both electric and magnetic fields.

Complete step by step solution:
The notion of an electric field is used to illustrate how a charge, or a group of charges, impacts the space surrounding it. The electric field E is comparable to g, which we used to refer to as gravity's acceleration but is actually the gravitational field. Everything we've learned about gravity and how masses react to gravity may help us comprehend how electric charges react to electric forces. The electric field is given by at a distance r from a point charge Q.
Electric field from a point charge : \[E\text{ }=\dfrac{\text{ }k\text{ }{{Q}^{2}}}{r}\]
A positive charge's electric field points away from the charge, whereas a negative charge's electric field points toward the charge. The electric field E, like the electric force, is a vector.
When the electric field at a specific position is known, the force that a charge q experiences when placed there is given by: F = qE
The force is in the same direction as the field if q is positive; if q is negative, the force is in the opposite direction.
You are now surrounded by a uniform gravitational field with a magnitude of \[9.8\text{ }m{{s}^{-2}}\] and a direction of straight down. Because of gravity, if you hurled a mass into the air, it would follow a parabolic path. You might do a projectile motion analysis to figure out when and where the object would land by splitting everything into x and y components. The vertical acceleration is g, while the horizontal acceleration is zero. We know this because a free-body diagram only displays mg working vertically, and Newton's second rule informs us that mg = ma, implying that a = g.
Charges in a uniform electric field can be used in the same way. A charge would likewise follow a parabolic route if it were thrown into a uniform electric field (the same magnitude and direction everywhere). We'll ignore gravity; the parabola is caused by the charge in the electric field experiencing a constant force. A projectile motion analysis may be used to predict when and where the charge will fall. In one direction, the acceleration is zero, whereas in the other, it is continuous. Drawing a free-body diagram (one force, \[F\text{ }=\text{ }qE\] ) and using Newton's second law will provide the acceleration value. The acceleration is \[g\text{ }=\text{ }\dfrac{qE}{m}\text{ }\]since \[qE\text{ }=\text{ }mg\].
For one electron q = E
Hence \[eE\text{ }=\text{ }mg\]
\[\therefore E\text{ }=\text{ }\dfrac{mg}{e}\]
Hence option D is correct.

Note:
On paper, an electric field may be seen by drawing lines of force, which indicate the magnitude and strength of the field. Field lines are another name for force lines. The direction of the field line at a point tells you what direction the force experienced by a charge will be if the charge is placed there. Field lines start on positive charges and end on negative charges, and the direction of the field line at a point tells you what direction the force experienced by a charge will be if the charge is placed there. If the charge is positive, the force will be in the same direction as the field; if the charge is negative, the force will be in the other direction.