Calculate the radius of a drop of water which remains suspended in the eart's electric field of 300 V`m^(-1)` when charged with one electronic charge. (One electronic charge=`1.6xx10^(-19)`C)

Answer

To calculate the radius of a drop of water that remains suspended in the Earth's electric field of 300 V/m when charged with one electronic charge, we can follow these steps:

### Step 1: Understand the forces acting on the water drop
When the water drop is suspended in the electric field, two forces act on it:
1. The gravitational force (weight) acting downwards, given by \( F_g = mg \).
2. The electric force acting upwards, given by \( F_e = qE \).

### Step 2: Set up the equilibrium condition
For the drop to remain suspended, the electric force must balance the gravitational force:
\[ F_e = F_g \]
This gives us the equation:
\[ qE = mg \]

### Step 3: Substitute known values
We know:
- The charge \( q \) on the drop is equal to the charge of one electron, \( q = 1.6 \times 10^{-19} \, \text{C} \).
- The electric field \( E \) is given as \( 300 \, \text{V/m} \).
- The acceleration due to gravity \( g \) is approximately \( 9.8 \, \text{m/s}^2 \).

Substituting these values into the equation:
\[ (1.6 \times 10^{-19}) \times (300) = mg \]

### Step 4: Express mass in terms of volume and density
The mass \( m \) of the water drop can be expressed in terms of its volume and density:
\[ m = \rho V \]
For a spherical drop, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the drop and \( \rho \) is the density of water, approximately \( 1000 \, \text{kg/m}^3 \).

### Step 5: Substitute mass into the equilibrium equation
Substituting \( m \) into the equation gives:
\[ (1.6 \times 10^{-19}) \times (300) = \rho \left( \frac{4}{3} \pi r^3 \right) g \]
This simplifies to:
\[ (1.6 \times 10^{-19}) \times (300) = (1000) \left( \frac{4}{3} \pi r^3 \right) (9.8) \]

### Step 6: Solve for \( r^3 \)
Rearranging the equation to solve for \( r^3 \):
\[ r^3 = \frac{(1.6 \times 10^{-19}) \times (300)}{(1000) \left( \frac{4}{3} \pi \right) (9.8)} \]

### Step 7: Calculate the numerical values
Calculating the left-hand side:
\[ 1.6 \times 300 = 480 \]
Now substituting this into the equation:
\[ r^3 = \frac{480 \times 10^{-19}}{(1000) \left( \frac{4}{3} \pi \right) (9.8)} \]

Calculating the denominator:
\[ \frac{4}{3} \pi \approx 4.18879 \]
Thus,
\[ 1000 \times 4.18879 \times 9.8 \approx 4105.6 \]
Now substituting back:
\[ r^3 = \frac{480 \times 10^{-19}}{4105.6} \]
Calculating this gives:
\[ r^3 \approx 1.167 \times 10^{-22} \, \text{m}^3 \]

### Step 8: Take the cube root to find \( r \)
Finally, taking the cube root:
\[ r \approx (1.167 \times 10^{-22})^{1/3} \approx 1.05 \times 10^{-7} \, \text{m} \]

### Final Answer
The radius of the drop of water is approximately \( 1.05 \times 10^{-7} \, \text{m} \).

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