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 \(20 \, \mu C\) and \(80 \, \mu C\) that are \(10 \, cm\) apart, we can follow these steps:

### Step 1: Understand the Setup
We have two charges:
- Charge \(Q_1 = 20 \, \mu C\) (located at point A)
- Charge \(Q_2 = 80 \, \mu C\) (located at point B)

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

### Step 2: Identify the Region for Zero Electric Field
Since both charges are positive, the electric fields due to both charges will repel. The electric field will be zero at a point where the magnitudes of the electric fields due to both charges are equal. This point will be located on the line joining the two charges, but since both are positive, it will be to the left of the \(20 \, \mu C\) charge.

### Step 3: Set Up the Electric Field Equations
Let \(x\) be the distance from the \(20 \, \mu C\) charge where the electric field is zero. The distance from the \(80 \, \mu C\) charge will then be \(10 \, cm - x\).

The electric field \(E_1\) due to \(Q_1\) at distance \(x\) is given by:
\[ E_1 = \frac{k \cdot |Q_1|}{x^2} \]
The electric field \(E_2\) due to \(Q_2\) at distance \(10 \, cm - x\) is given by:
\[ E_2 = \frac{k \cdot |Q_2|}{(10 - x)^2} \]

### Step 4: Set the Electric Fields Equal
For the electric field to be zero:
\[ E_1 = E_2 \]
This gives us:
\[ \frac{k \cdot 20 \times 10^{-6}}{x^2} = \frac{k \cdot 80 \times 10^{-6}}{(10 - x)^2} \]

### Step 5: Simplify the Equation
We can cancel \(k\) from both sides:
\[ \frac{20 \times 10^{-6}}{x^2} = \frac{80 \times 10^{-6}}{(10 - x)^2} \]
This simplifies to:
\[ \frac{20}{x^2} = \frac{80}{(10 - x)^2} \]
Cross-multiplying gives:
\[ 20(10 - x)^2 = 80x^2 \]
Dividing both sides by 20:
\[ (10 - x)^2 = 4x^2 \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ 100 - 20x + x^2 = 4x^2 \]
Rearranging gives:
\[ 0 = 3x^2 + 20x - 100 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 3\), \(b = 20\), and \(c = -100\):
\[ x = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 3 \cdot (-100)}}{2 \cdot 3} \]
\[ x = \frac{-20 \pm \sqrt{400 + 1200}}{6} \]
\[ x = \frac{-20 \pm \sqrt{1600}}{6} \]
\[ x = \frac{-20 \pm 40}{6} \]
This gives two solutions:
1. \(x = \frac{20}{6} = \frac{10}{3} \, cm \approx 3.33 \, cm\) (valid)
2. \(x = \frac{-60}{6} = -10 \, cm\) (not valid as distance cannot be negative)

### Conclusion
The electric field strength will be zero at a distance of \(3.33 \, cm\) from the \(20 \, \mu C\) charge.