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Question: an oil drop having 12 excess electrons is held stationary under a uniform electric field of 2.55*10...

an oil drop having 12 excess electrons is held stationary under a uniform electric field of 2.55*10^4 N C^-1 the density of the oil is 1.26 g cm^-3 , estimate the radius of the drop , given , g=9.81 m s^-2 , e= 1.6 *10^-19 C

Answer

0.982 * 10^-6 m

Explanation

Solution

The oil drop is held stationary under the influence of a uniform electric field. This means the net force on the oil drop is zero. The forces acting on the drop are the gravitational force (weight) acting downwards and the electric force acting upwards (since the drop has excess electrons and the field is oriented to counteract gravity).

The gravitational force is given by Fg=mgF_g = mg, where mm is the mass of the oil drop and gg is the acceleration due to gravity.
The mass of the oil drop can be expressed in terms of its density ρ\rho and volume VV. Assuming the oil drop is spherical with radius rr, its volume is V=43πr3V = \frac{4}{3}\pi r^3. So, m=ρV=ρ(43πr3)m = \rho V = \rho \left(\frac{4}{3}\pi r^3\right).
Thus, Fg=ρ(43πr3)gF_g = \rho \left(\frac{4}{3}\pi r^3\right) g.

The electric force is given by Fe=qEF_e = qE, where qq is the charge of the oil drop and EE is the electric field strength.
The oil drop has 12 excess electrons, so its charge is q=n×eq = n \times e, where n=12n = 12 is the number of excess electrons and e=1.6×10−19e = 1.6 \times 10^{-19} C is the charge of an electron.
Thus, q=12×1.6×10−19q = 12 \times 1.6 \times 10^{-19} C.

Since the oil drop is stationary, the forces are balanced: Fe=FgF_e = F_g.
qE=mgqE = mg
(n×e)E=ρ(43πr3)g(n \times e) E = \rho \left(\frac{4}{3}\pi r^3\right) g

We need to find the radius rr. Rearranging the equation to solve for r3r^3:
r3=(n×e)Eρ(43πg)r^3 = \frac{(n \times e) E}{\rho \left(\frac{4}{3}\pi g\right)}
r3=3(n×e)E4πρgr^3 = \frac{3 (n \times e) E}{4 \pi \rho g}

Now, substitute the given values:
n=12n = 12
e=1.6×10−19e = 1.6 \times 10^{-19} C
E=2.55×104E = 2.55 \times 10^4 N C−1^{-1}
ρ=1.26\rho = 1.26 g cm−3^{-3}
g=9.81g = 9.81 m s−2^{-2}

First, convert the density to kg m−3^{-3}:
ρ=1.26 g cm−3=1.26×10−3 kg(10−2 m)3=1.26×10−3 kg10−6 m3=1.26×103 kg m−3\rho = 1.26 \text{ g cm}^{-3} = 1.26 \times \frac{10^{-3} \text{ kg}}{(10^{-2} \text{ m})^3} = 1.26 \times \frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} = 1.26 \times 10^3 \text{ kg m}^{-3}

Calculate the total charge qq:
q=12×1.6×10−19q = 12 \times 1.6 \times 10^{-19} C =19.2×10−19= 19.2 \times 10^{-19} C

Substitute all values into the equation for r3r^3:
r3=3×(19.2×10−19 C)×(2.55×104 N C−1)4×π×(1.26×103 kg m−3)×(9.81 m s−2)r^3 = \frac{3 \times (19.2 \times 10^{-19} \text{ C}) \times (2.55 \times 10^4 \text{ N C}^{-1})}{4 \times \pi \times (1.26 \times 10^3 \text{ kg m}^{-3}) \times (9.81 \text{ m s}^{-2})}
r3=3×19.2×2.55×10−19+44×π×1.26×9.81×103 m3r^3 = \frac{3 \times 19.2 \times 2.55 \times 10^{-19+4}}{4 \times \pi \times 1.26 \times 9.81 \times 10^3} \text{ m}^3
r3=146.88×10−154π×12.3606×103 m3r^3 = \frac{146.88 \times 10^{-15}}{4 \pi \times 12.3606 \times 10^3} \text{ m}^3 (Using π≈3.14159\pi \approx 3.14159)
r3=146.88×10−15155.208×103 m3r^3 = \frac{146.88 \times 10^{-15}}{155.208 \times 10^3} \text{ m}^3
r3=146.88155.208×10−15−3 m3r^3 = \frac{146.88}{155.208} \times 10^{-15-3} \text{ m}^3
r3≈0.94634×10−18 m3r^3 \approx 0.94634 \times 10^{-18} \text{ m}^3

Now, take the cube root to find rr:
r=(0.94634×10−18)1/3 mr = (0.94634 \times 10^{-18})^{1/3} \text{ m}
r=(0.94634)1/3×(10−18)1/3 mr = (0.94634)^{1/3} \times (10^{-18})^{1/3} \text{ m}
r≈0.9818×10−6 mr \approx 0.9818 \times 10^{-6} \text{ m}

Rounding to three significant figures (based on the precision of the given values like 2.55, 1.26, 9.81):
r≈0.982×10−6 mr \approx 0.982 \times 10^{-6} \text{ m}