An object having a net charge of 18.2 μC is placed in a uniform electric field of 790 N/C directed vertically upward, i.e. in the direction opposite gravity. What is the mass, in grams, of this object if it "floats" in the field? (Assume g = 9.80 m/s2.)
Google Millikan oil drop experiment
for example:
http://ffden-2.phys.uaf.edu/212_fall2003.web.dir/Ryan_McAllister/Slide3.htm
To find the mass of the object, we can use the concept of electrostatic force and gravitational force. Since the object is "floating," it means that the electrostatic force is equal to the gravitational force.
The electrostatic force can be calculated using the formula:
F_e = qE
where F_e is the electrostatic force, q is the charge of the object, and E is the electric field.
The gravitational force can be calculated using the formula:
F_g = mg
where F_g is the gravitational force, m is the mass of the object, and g is the acceleration due to gravity.
Since F_e = F_g, we can equate the two forces:
qE = mg
Rearranging the equation, we can solve for the mass of the object:
m = (qE) / g
Substituting the given values:
q = 18.2 μC = 18.2 * 10^(-6) C
E = 790 N/C
g = 9.80 m/s^2
m = (18.2 * 10^(-6) C * 790 N/C) / 9.80 m/s^2
m = 1.468056 * 10^(-3) kg
Now, to convert the mass from kg to grams, we can multiply by 1000:
m = (1.468056 * 10^(-3) kg) * 1000
m ≈ 1.468056 g
Therefore, the mass of the object is approximately 1.468 grams.