Two point charges of `20 mu C` and `80 muC` are 10 cm apart where will the electric field strength be zero on the line joining the charges from `20 mu C` charge

Options

A. 0.1 m
B. 0.04 m
C. 0.033 m
D. 0.33 m

Answer

To find the point where the electric field strength is zero on the line joining two point charges of `20 µC` and `80 µC` that are `10 cm` apart, we can follow these steps:

### Step 1: Understand the Setup
We have two charges:
- Charge \( Q_1 = 20 \, \mu C = 20 \times 10^{-6} \, C \)
- Charge \( Q_2 = 80 \, \mu C = 80 \times 10^{-6} \, C \)

The distance between the charges is \( d = 10 \, cm = 0.1 \, m \).

### Step 2: Identify the Point of Interest
We need to find a point \( P \) on the line joining the two charges where the electric field strength \( E \) is zero. This point will be closer to the smaller charge \( Q_1 \) since it has a smaller magnitude.

### Step 3: Set Up the Electric Field Equations
Let \( x \) be the distance from charge \( Q_1 \) to the point \( P \). Therefore, the distance from \( Q_2 \) to point \( P \) will be \( (0.1 - x) \).

The electric field due to \( Q_1 \) at point \( P \) is given by:
\[ E_1 = \frac{k \cdot Q_1}{x^2} \]
The electric field due to \( Q_2 \) at point \( P \) is given by:
\[ E_2 = \frac{k \cdot Q_2}{(0.1 - x)^2} \]

### Step 4: Set the Electric Fields Equal
For the electric field to be zero at point \( P \), the magnitudes of \( E_1 \) and \( E_2 \) must be equal:
\[ E_1 = E_2 \]
\[ \frac{k \cdot Q_1}{x^2} = \frac{k \cdot Q_2}{(0.1 - x)^2} \]

### Step 5: Cancel Out Constants
Since \( k \) is a constant, we can cancel it out from both sides:
\[ \frac{Q_1}{x^2} = \frac{Q_2}{(0.1 - x)^2} \]

### Step 6: Substitute the Values
Substituting the values of \( Q_1 \) and \( Q_2 \):
\[ \frac{20 \times 10^{-6}}{x^2} = \frac{80 \times 10^{-6}}{(0.1 - x)^2} \]

### Step 7: Cross Multiply
Cross multiplying gives:
\[ 20 \times 10^{-6} \cdot (0.1 - x)^2 = 80 \times 10^{-6} \cdot x^2 \]
Dividing both sides by \( 10^{-6} \):
\[ 20 \cdot (0.1 - x)^2 = 80 \cdot x^2 \]

### Step 8: Simplify the Equation
Dividing both sides by 20:
\[ (0.1 - x)^2 = 4x^2 \]

### Step 9: Expand and Rearrange
Expanding the left side:
\[ 0.01 - 0.2x + x^2 = 4x^2 \]
Rearranging gives:
\[ 0 = 3x^2 + 0.2x - 0.01 \]

### Step 10: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = 0.2 \), and \( c = -0.01 \).
\[ x = \frac{-0.2 \pm \sqrt{(0.2)^2 - 4 \cdot 3 \cdot (-0.01)}}{2 \cdot 3} \]
\[ x = \frac{-0.2 \pm \sqrt{0.04 + 0.12}}{6} \]
\[ x = \frac{-0.2 \pm \sqrt{0.16}}{6} \]
\[ x = \frac{-0.2 \pm 0.4}{6} \]

Calculating the two possible values:
1. \( x = \frac{0.2}{6} = 0.0333 \, m \)
2. \( x = \frac{-0.6}{6} \) (not a valid solution since distance cannot be negative)

### Final Answer
Thus, the point where the electric field strength is zero is:
\[ x = 0.0333 \, m \text{ or } 3.33 \, cm \text{ from the } 20 \, \mu C \text{ charge.} \]