Two uniformly long charged wires with linear densities `lambda` and `3lambda` are placed along X and Y axis respectively. Determined the slope of electric field at any point on the l ine `y=sqrt3x`

Options

A. `3sqrt3`
B. `(sqrt3)/(3sqrt2)`
C. `(1)/(3sqrt3)`
D. `sqrt3`

Answer

To solve the problem, we need to find the slope of the electric field at any point on the line \( y = \sqrt{3}x \) due to two charged wires. One wire is placed along the x-axis with linear charge density \( \lambda \), and the other wire is along the y-axis with linear charge density \( 3\lambda \).

### Step-by-Step Solution:

1. **Identify the Electric Field due to Each Wire**:
- For an infinite line of charge, the electric field \( E \) at a perpendicular distance \( d \) from the wire with linear charge density \( \lambda \) is given by:
\[ E = \frac{2k\lambda}{d} \]
where \( k \) is Coulomb's constant.

2. **Electric Field from the Wire along the X-axis**:
- The wire along the x-axis has a linear charge density \( \lambda \).
- At any point \( P \) on the line \( y = \sqrt{3}x \), the perpendicular distance from the x-axis (y = 0) to point \( P \) is \( y = \sqrt{3}x \).
- Therefore, the electric field \( E_x \) due to this wire at point \( P \) is directed upwards (positive y-direction):
\[ E_x = \frac{2k\lambda}{\sqrt{3}x} \]

3. **Electric Field from the Wire along the Y-axis**:
- The wire along the y-axis has a linear charge density \( 3\lambda \).
- The perpendicular distance from the y-axis (x = 0) to point \( P \) is \( x \).
- Therefore, the electric field \( E_y \) due to this wire at point \( P \) is directed away from the wire (positive x-direction):
\[ E_y = \frac{2k(3\lambda)}{x} = \frac{6k\lambda}{x} \]

4. **Components of the Electric Field**:
- The electric field components at point \( P \) are:
- \( E_x \) in the y-direction: \( \frac{2k\lambda}{\sqrt{3}x} \)
- \( E_y \) in the x-direction: \( \frac{6k\lambda}{x} \)

5. **Finding the Slope of the Electric Field**:
- The slope \( m \) of the electric field vector \( \vec{E} \) at point \( P \) can be found using the ratio of the y-component to the x-component:
\[ m = \frac{E_x}{E_y} = \frac{\frac{2k\lambda}{\sqrt{3}x}}{\frac{6k\lambda}{x}} \]
- Simplifying this expression:
\[ m = \frac{2k\lambda}{\sqrt{3}x} \cdot \frac{x}{6k\lambda} = \frac{2}{6\sqrt{3}} = \frac{1}{3\sqrt{3}} \]

### Final Answer:
The slope of the electric field at any point on the line \( y = \sqrt{3}x \) is:
\[ \frac{1}{3\sqrt{3}} \]