Question

In 1909, Robert Millikan was the first to find the charge of an electron in his now famous oil drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended upward force qE just equaled mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of $1.6 \times {10^{ - 19}}\;{\text{C}}$ the charge on the electron. For this, he won the Nobel prize. If a drop of mass $1.08 \times {10^{ - 14}}\;{\text{kg}}$ remains stationary in an electric field of$1.68 \times {10^5}\;{\text{N}}{{\text{C}}^{ - 1}}$, the charge of this drop is:
A. $6.40 \times {10^{ - 19}}\;{\text{C}}$
B. $3.2 \times {10^{ - 19}}\;{\text{C}}$
C. $1.6 \times {10^{ - 19}}\;{\text{C}}$
D. $4.8 \times {10^{ - 19}}\;{\text{C}}$

Answer 1

Hint:The above problem is based on the Millikan’s oil drop experiment. In this experiment the charge of the electron is found by balancing the force of gravity by the electric force opposite to the force of gravity. The electric force and force due to gravity become equal between the parallel charged plates.

Complete step by step answer:
Given:
The mass of the oil drop is $m = 1.08 \times {10^{ - 14}}\;{\text{kg}}$
The electric field strength is \[E = 1.68 \times {10^5}\;{\text{N}}{{\text{C}}^{ - 1}}\]
The expression to calculate the electric force on the drop is given as,
${F_e} = qE......\left( 1 \right)$
Here, q is the charge of the oil drop.
The expression to calculate the force of gravity on the drop is given as,
${F_g} = mg......\left( 2 \right)$
Here, g is the gravitational acceleration and its value is $10\;{\text{m}}/{{\text{s}}^2}$.
As the electric force on the oil drop balances the force of gravity of the oil drop, so equate the expression (1) and expression (2).
${F_e} = {F_g}$
$\Rightarrow qE = mg$
$\Rightarrow q = \dfrac{{mg}}{E}......\left( 3 \right)$
Substitute $1.08 \times {10^{ - 14}}\;{\text{kg}}$for m, $10\;{\text{m}}/{{\text{s}}^2}$for g and $1.68 \times {10^5}\;{\text{N}}{{\text{C}}^{ - 1}}$for E in the expression (30 to find the charge on the oil drop.
$q = \dfrac{{\left( {1.08 \times {{10}^{ - 14}}\;{\text{kg}}} \right)\left( {10\;{\text{m}}/{{\text{s}}^2}} \right)}}{{1.68 \times {{10}^5}\;{\text{N}}{{\text{C}}^{ - 1}}}}$
$\Rightarrow q = 6.42 \times {10^{ - 19}}\;{\text{C}}$
$\therefore q \approx 6.40 \times {10^{ - 19}}\;{\text{C}}$

Thus, the charge of the oil drop is $6.40 \times {10^{ - 19}}\;{\text{C}}$and the option (A) is the correct answer.

Additional Information:
The electric field is the region around the charged particle in which the other particles experience the effect of the charged particle. This region varies with the charge of the particle and its distance from the other particles.

Note: Take the value of gravitational acceleration as $10\;{\text{m}}/{{\text{s}}^2}$ in place of $9.8\;{\text{m}}/{{\text{s}}^2}$ to calculate the correct value of the charge in the above problem. Always remember that charge of any object must be multiple of unit charge.