Question

Robert Millikan was an American physicist who in 1909 performed a classic experiment involving charged oil drops. Millikan adjusted the voltage applied between two metal plates in order to suspend charged drops in the resulting electric field. For a suspended drop, the electrostatic force directed up exactly balances out the weight of the drop. Millikan ultimately determined the charge on an electron and that charge is quantized. This means that charge comes in increments, discrete amounts, rather than any amount.A NEGATIVELY-charged oil drop is suspended between two plates of opposite charge by an electric field of 402 newtons per coulomb. The drop is determined to have a mass of 3.28 x 10-17 kilograms. What is the charge on the drop? Include units in your answer.

Answer 1

The charge on the oil drop is calculated using the formula q = mg/E, yielding a charge of 8.03 x 10^-19 C. This value represents a quantized multiple of the elementary charge on an electron.

Step-by-step explanation:

The student is asking about a physics problem related to Robert Millikan's oil drop experiment, which was conducted to determine the charge of an electron. In the problem, the electrostatic force exerted on a negatively-charged oil drop with a known mass is balanced against the drop's weight when it is suspended in an electric field. The charge on the oil drop can be found using the formula F = Eq = mg, where F is the electrostatic force, E is the electric field strength, q is the charge on the oil drop, m is the mass of the oil drop, and g is the acceleration due to gravity, which is approximately 9.8 m/s2.

To find the charge on the oil drop (q), you can rearrange the formula to q = mg/E. Plugging in the given values, q = (3.28 x 10-17 kg) (9.8 m/s2) / (402 N/C), we get q = (3.28 x 10-17 kg) (9.8 m/s2) / (402 N/C) = q = 8.03 x 10-19 C. Therefore, the charge on the oil drop is 8.03 x 10-19 C.

This measured charge would be quantized in multiples of the elementary charge for an electron, which is -1.6 x 10-19 C.

Answer 2

Given

E: Electric field

E = 402 N/C

m: mass

m = 3.28 x 10-17 kg

Procedure

Let q be the amount of the charge that the drop carries.

The coulomb force balances the gravitational force acting on the drop at equilibrium.

`\begin{gathered} qE=mg \\ q=(mg)/(E) \\ q=\frac{3.28*10^(-17)kg*9.8m/s^2}{402\text{ N/C}} \\ q=8.09*10^(-19)C \end{gathered}`

The answer would be 8.09x10^-19 C