Question

A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electrical field of strength \(\frac{81}{7}\pi\) x 10⁵ V/m.When the field is switched off, the drop is observed to fall with terminal velocity 2 x 10^-3 m/s. Here g = 9.8 m/s², viscosity of air is 1.8 x 10^-5 N/m² and the density of the oil is 900 kg/m³. The magnitude of 'q' is 

• A. 1.6 x 10^-19 C
• B. 0.8 x 10^-19 C
• C. 3.2 x 10^-19 C
• D. 8 x 10^-19 C

Answer 1

The Correct Option is D

Step 1: Equating Gravitational Force and Drag Force 

The gravitational force acting on the drop is given by:

\( F_{\text{gravity}} = m \cdot g \)

The drag force due to viscosity is given by Stokes' law:

\( F_{\text{drag}} = 6 \pi \eta r v \)

At terminal velocity, these two forces are balanced:

\( m \cdot g = 6 \pi \eta r v \)

Step 2: Solving for Mass

Rearranging the equation for \( m \):

\( m = \frac{6 \pi \eta r v}{g} \)

The mass of the drop can also be expressed in terms of its volume and density:

\( m = \frac{4}{3} \pi r^3 \rho \)

Substitute this expression for \( m \) into the equation:

\( \frac{4}{3} \pi r^3 \rho \cdot g = 6 \pi \eta r v \)

Step 3: Solving for the Radius \( r \)

Rearrange the equation to solve for \( r^2 \):

\( r^2 = \frac{9 \eta v}{2 \rho g} \)

Step 4: Substituting the Given Values

Given values:

• Viscosity \( \eta = 1.8 \times 10^{-5} \, \text{N/m}^2 \)
• Velocity \( v = 2 \times 10^{-3} \, \text{m/s} \)
• Density \( \rho = 900 \, \text{kg/m}^3 \)
• Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)

Substitute these values into the equation:

\( r^2 = \frac{9 \times 1.8 \times 10^{-5} \times 2 \times 10^{-3}}{2 \times 900 \times 9.8} \

Simplifying:

\( r^2 = \frac{9 \times 10^{-8}}{17640} \approx 5.10 \times 10^{-13} \)

Taking the square root of both sides:

\( r \approx 7.14 \times 10^{-7} \, \text{m} \)

Step 5: Finding the Charge \( q \)

The electric force acting on the oil drop is given by:

\( F_{\text{electric}} = q \cdot E \)

The electric field strength is given as:

\( E = \frac{81 \pi}{7 \times 10^3} \, \text{V/m} \)

The gravitational force can also be expressed as:

\( F_{\text{electric}} = \frac{4}{3} \pi r^3 \rho \cdot g \)

Equating the two expressions for the electric force:

\( q \cdot E = \frac{4}{3} \pi r^3 \rho \cdot g \)

Solving for \( q \):

\( q = \frac{\frac{4}{3} \pi r^3 \rho g}{E} \)

Substitute the known values:

\( q = \frac{\frac{4}{3} \pi \cdot (7.14 \times 10^{-7})^3 \cdot 900 \cdot 9.8}{\frac{81 \pi}{7 \times 10^3}} \)

Simplifying:

\( q \approx 8 \times 10^{-19} \, \text{C} \)

Therefore, the magnitude of the charge \( q \) is: \( 8 \times 10^{-19} \, \text{C} \) (option D).

Answer 2

Given the following equations:

\( qE = mg \quad \text{...(i)} \)

\( 6 \pi \eta r v = mg \) 

Also, we have:

\( \frac{3}{4} \pi r^3 \rho g = mg \quad \text{...(ii)} \)

Solving for \( r \), we get:

\( r = \left( \frac{4 \pi \rho g}{3 mg} \right)^{1/3} \quad \text{...(iii)} \)

Substituting the value of \( r \) in Equation (ii), we get:

\( 6 \pi \eta v \left( \frac{4 \pi \rho g}{3 mg} \right)^{1/3} = mg \)

Now, simplifying this expression:

\( (6 \pi \eta v)^3 \left( \frac{4 \pi \rho g}{3 mg} \right) = (mg)^3 \)

Again, substituting \( mg = qE \), we get:

\( (qE)^2 = \left( \frac{4 \pi \rho g}{3} \right) \cdot (6 \pi \eta v)^3 \)

Thus, solving for \( qE \), we get:

\( qE = \left( \frac{4 \pi \rho g}{3} \right)^{1/2} \cdot (6 \pi \eta v)^{3/2} \)

Finally, we find \( q \) as:

\( q = E \left( \frac{4 \pi \rho g}{3} \right)^{1/2} \cdot (6 \pi \eta v)^{3/2} \)

Substituting the values:

\( q = \frac{81 \pi \times 10^5}{\sqrt{4 \pi \times 900 \times 9.8}} \times 3 \times 216 \pi \times (1.8 \times 10^{-5} \times 2 \times 10^{-3})^3 \)

Therefore, the value of \( q \) is: \( 8.0 \times 10^{-19} \, \text{C} \).