To find the point along the line joining the masses where the gravitational field due to them is zero, we can follow these steps:
### Step 1: Understand the Setup
We have two masses:
- \( m_1 = 1 \, \text{kg} \)
- \( m_2 = 4 \, \text{kg} \)
They are separated by a distance of \( d = 6 \, \text{cm} \).
### Step 2: Define the Point of Interest
Let \( P \) be the point where the gravitational field is zero. Let \( x \) be the distance from the 1 kg mass to point \( P \). Therefore, the distance from the 4 kg mass to point \( P \) will be \( (6 - x) \, \text{cm} \).
### Step 3: Set Up the Gravitational Field Equations
The gravitational field \( E \) due to a mass \( m \) at a distance \( r \) is given by:
\[
E = \frac{Gm}{r^2}
\]
Where \( G \) is the gravitational constant.
At point \( P \), the gravitational fields due to both masses must be equal in magnitude but opposite in direction:
\[
E_1 = E_2
\]
Thus,
\[
\frac{G m_1}{(6 - x)^2} = \frac{G m_2}{x^2}
\]
Since \( G \) appears on both sides, we can cancel it out:
\[
\frac{m_1}{(6 - x)^2} = \frac{m_2}{x^2}
\]
### Step 4: Substitute the Masses
Substituting \( m_1 = 1 \, \text{kg} \) and \( m_2 = 4 \, \text{kg} \):
\[
\frac{1}{(6 - x)^2} = \frac{4}{x^2}
\]
### Step 5: Cross-Multiply
Cross-multiplying gives:
\[
1 \cdot x^2 = 4 \cdot (6 - x)^2
\]
Expanding the right side:
\[
x^2 = 4(36 - 12x + x^2)
\]
\[
x^2 = 144 - 48x + 4x^2
\]
### Step 6: Rearrange the Equation
Rearranging gives:
\[
0 = 4x^2 - x^2 - 48x + 144
\]
\[
0 = 3x^2 - 48x + 144
\]
### Step 7: Simplify the Quadratic Equation
Dividing the entire equation by 3:
\[
0 = x^2 - 16x + 48
\]
### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -16, c = 48 \):
\[
x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1}
\]
\[
x = \frac{16 \pm \sqrt{256 - 192}}{2}
\]
\[
x = \frac{16 \pm \sqrt{64}}{2}
\]
\[
x = \frac{16 \pm 8}{2}
\]
Calculating the two possible values:
1. \( x = \frac{24}{2} = 12 \, \text{cm} \)
2. \( x = \frac{8}{2} = 4 \, \text{cm} \)
### Step 9: Determine the Valid Solution
Since \( x = 12 \, \text{cm} \) is outside the distance between the two masses, we discard it. Thus, the valid solution is:
\[
x = 4 \, \text{cm}
\]
### Conclusion
The point along the line joining the masses where the gravitational field due to them is zero is at a distance of \( 4 \, \text{cm} \) from the \( 1 \, \text{kg} \) mass.
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