Two charges 4q and q are fixed at points (0,9) and (12, 0) respectively on the X-Y plane. Find the coordinates of the point where the electric field strength is zero.

Answer

To find the coordinates of the point where the electric field strength is zero due to the two charges \(4q\) and \(q\) located at points \((0, 9)\) and \((12, 0)\) respectively, we can follow these steps:

### Step 1: Understand the Setup
We have two charges:
- Charge \(4q\) is located at point \(A(0, 9)\).
- Charge \(q\) is located at point \(B(12, 0)\).

### Step 2: Calculate the Distance Between the Charges
Using the distance formula, the distance \(d\) between points \(A\) and \(B\) can be calculated as:
\[ d = \sqrt{(12 - 0)^2 + (0 - 9)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \]

### Step 3: Set Up the Electric Field Equations
Let \(P\) be the point where the electric field is zero, located at a distance \(x\) from charge \(4q\) and \(15 - x\) from charge \(q\). The electric field \(E\) due to a point charge is given by:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where \(k\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance from the charge.

The electric field due to \(4q\) at point \(P\) is:
\[ E_1 = \frac{k \cdot 4q}{x^2} \]
The electric field due to \(q\) at point \(P\) is:
\[ E_2 = \frac{k \cdot q}{(15 - x)^2} \]

### Step 4: Set the Electric Fields Equal
Since the electric field is zero at point \(P\), we set \(E_1\) equal to \(E_2\):
\[ \frac{k \cdot 4q}{x^2} = \frac{k \cdot q}{(15 - x)^2} \]
We can cancel \(k\) and \(q\) (assuming \(q \neq 0\)):
\[ \frac{4}{x^2} = \frac{1}{(15 - x)^2} \]

### Step 5: Cross Multiply and Simplify
Cross-multiplying gives:
\[ 4(15 - x)^2 = x^2 \]
Expanding the left side:
\[ 4(225 - 30x + x^2) = x^2 \]
This simplifies to:
\[ 900 - 120x + 4x^2 = x^2 \]
Rearranging gives:
\[ 3x^2 - 120x + 900 = 0 \]

### Step 6: Solve the Quadratic Equation
Dividing the entire equation by 3:
\[ x^2 - 40x + 300 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 1 \cdot 300}}{2 \cdot 1} = \frac{40 \pm \sqrt{1600 - 1200}}{2} = \frac{40 \pm \sqrt{400}}{2} = \frac{40 \pm 20}{2} \]
This results in:
\[ x = \frac{60}{2} = 30 \quad \text{or} \quad x = \frac{20}{2} = 10 \]

### Step 7: Determine the Valid \(x\)
Since the total distance between the charges is 15, \(x = 30\) is not valid. Therefore, we take \(x = 10\).

### Step 8: Find the Coordinates of Point \(P\)
Now, we need to find the coordinates of point \(P\):
- The distance from \(4q\) is \(x = 10\).
- The distance from \(q\) is \(15 - x = 5\).

Using the section formula, we find the coordinates of point \(P\):
\[ x_P = \frac{m_1 \cdot x_2 + m_2 \cdot x_1}{m_1 + m_2} = \frac{2 \cdot 12 + 1 \cdot 0}{2 + 1} = \frac{24}{3} = 8 \]
\[ y_P = \frac{m_1 \cdot y_2 + m_2 \cdot y_1}{m_1 + m_2} = \frac{2 \cdot 0 + 1 \cdot 9}{2 + 1} = \frac{9}{3} = 3 \]

### Final Coordinates
Thus, the coordinates of the point where the electric field strength is zero are:
\[ \text{Coordinates of } P = (8, 3) \]

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