A drop of `10^(-6)` kg water carries `10^(-6) C` charge. What electric field should be applied to balance its weight (assume `g = 10 ms^(-2)`)

Options

A. 10 V/m upward
B. 10 V/m downward
C. 0.1 V/m downward
D. 0.1 V/m upward

Answer

To solve the problem of determining the electric field required to balance the weight of a water drop carrying a charge, we can follow these steps:

### Step 1: Identify the forces acting on the drop
The drop experiences two main forces:
1. The gravitational force (weight) acting downwards, given by \( F_g = mg \).
2. The electrostatic force acting upwards, given by \( F_e = qE \), where \( E \) is the electric field strength.

### Step 2: Write down the expressions for the forces
- The gravitational force can be calculated using the formula:
\[ F_g = mg \]
- The electrostatic force can be calculated using the formula:
\[ F_e = qE \]

### Step 3: Set the forces equal for balance
For the drop to be in equilibrium (stationary), the upward electrostatic force must balance the downward gravitational force:
\[ F_e = F_g \]
This can be expressed as:
\[ qE = mg \]

### Step 4: Solve for the electric field \( E \)
Rearranging the equation gives:
\[ E = \frac{mg}{q} \]

### Step 5: Substitute the known values
Given:
- Mass of the drop, \( m = 10^{-6} \) kg
- Charge on the drop, \( q = 10^{-6} \) C
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

Substituting these values into the equation:
\[ E = \frac{(10^{-6} \, \text{kg})(10 \, \text{m/s}^2)}{10^{-6} \, \text{C}} \]

### Step 6: Calculate \( E \)
\[ E = \frac{10^{-5} \, \text{kg m/s}^2}{10^{-6} \, \text{C}} = 10 \, \text{V/m} \]

### Step 7: Determine the direction of the electric field
Since the charge on the drop is positive, the electric field must be directed upwards to exert an upward force that balances the downward weight of the drop.

### Final Answer
The electric field required to balance the weight of the drop is:
\[ E = 10 \, \text{V/m} \text{ (upward)} \]

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