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Step-by-Step Solution
Step 1: Understand the Physical Situation
The oil drop is held stationary in a uniform electric field. This means the upward electric force on the charged drop balances the downward gravitational force.
Step 2: Write the Force Balance
Because the drop is in equilibrium (stationary), the electric force equals the weight of the oil drop:
$ q E = mg $
where:
$q$ = charge on the oil drop (in coulombs),
$E$ = electric field strength (in V/m),
$m$ = mass of the oil drop (in kg),
$g$ = acceleration due to gravity (in m/s2).
Step 3: Determine the Mass of the Oil Drop
(a) Convert the given radius to SI units. The radius is given as 2 mm:
$ r = 2 \times 10^{-3} \text{ m} $.
(b) Compute the volume of the spherical drop:
$ V = \frac{4}{3} \pi r^3
= \frac{4}{3} \pi (2 \times 10^{-3})^3 \text{ m}^3
$.
(c) Convert the density to SI units. The given density is $3 \text{ g/cm}^3$. Since
$1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3,$
we have
$3 \text{ g/cm}^3 = 3000 \text{ kg/m}^3.$
(d) Calculate the mass:
$ m = \rho \times V = 3000 \times \frac{4}{3} \pi (2 \times 10^{-3})^3 \text{ kg}.$
Step 4: Express $q$ in Terms of $m, E, \text{ and } g$
From the equilibrium condition $q E = mg,$ solve for $q$:
$ q = \frac{mg}{E}.$
Step 5: Calculate the Number of Excess Electrons
The charge of one electron is
$ e = 1.6 \times 10^{-19} \text{ C}.$
The number of excess electrons $N$ on the drop is:
$ N = \frac{q}{e} = \frac{mg}{E \times e}.
$
Plug in the values:
$m = 3000 \times \frac{4}{3} \pi (2 \times 10^{-3})^3 \text{ kg},$
$g = 9.81 \text{ m/s}^2,$
$E = 3.55 \times 10^5 \text{ V/m},$
$e = 1.6 \times 10^{-19} \text{ C}.$
After simplification, the result is found to be approximately
$1.73 \times 10^{10}.$
Step 6: State the Final Answer
Hence, the number of excess electrons on the oil drop is:
$1.73 \times 10^{10}$
