If an oil drop of weight `3.2xx10^(-13) N` is balanced in an electric field of `5xx10^(5) Vm^(-1)`, find the charge on the oil drop.

Answer

To solve the problem of finding the charge on the oil drop, we can follow these steps:

### Step 1: Understand the Forces Acting on the Oil Drop
The oil drop is subjected to two main forces:
1. The gravitational force (weight) acting downwards, denoted as \( F_g \).
2. The electrostatic force acting upwards due to the electric field, denoted as \( F_e \).

### Step 2: Set Up the Equation
Since the oil drop is balanced in the electric field, the upward electrostatic force equals the downward gravitational force:
\[ F_e = F_g \]

### Step 3: Write the Expressions for the Forces
The gravitational force can be expressed as:
\[ F_g = mg \]
where \( m \) is the mass of the oil drop and \( g \) is the acceleration due to gravity.

The electrostatic force can be expressed as:
\[ F_e = Q \cdot E \]
where \( Q \) is the charge on the oil drop and \( E \) is the electric field strength.

### Step 4: Substitute the Known Values
From the problem statement:
- The weight of the oil drop \( F_g = 3.2 \times 10^{-13} \, \text{N} \)
- The electric field strength \( E = 5 \times 10^{5} \, \text{V/m} \)

Setting the forces equal gives:
\[ Q \cdot E = F_g \]

### Step 5: Solve for the Charge \( Q \)
Rearranging the equation to solve for \( Q \):
\[ Q = \frac{F_g}{E} \]

Substituting the known values:
\[ Q = \frac{3.2 \times 10^{-13} \, \text{N}}{5 \times 10^{5} \, \text{V/m}} \]

### Step 6: Calculate the Charge
Now, perform the calculation:
\[ Q = \frac{3.2}{5} \times 10^{-13} \times 10^{-5} = 0.64 \times 10^{-18} \, \text{C} \]

### Final Answer
Thus, the charge on the oil drop is:
\[ Q = 0.64 \times 10^{-18} \, \text{C} \]

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