The electric potential varies in space according to the relation `V = 3x + 4y`. A particle of mass 10 kg starts from rest from point `(2, 3.2)` m under the influence of this field. Find the velocityr,f the particle when it crosses the x-axis. The charge on the particle is `+1muC`. Assume V (x, y)are in SI units.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the Electric Field
The electric potential \( V \) is given by the equation:
\[ V = 3x + 4y \]
The electric field \( \mathbf{E} \) can be found using the relation:
\[ \mathbf{E} = -\nabla V \]
Calculating the gradient:
\[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) = (3, 4) \]
Thus, the electric field is:
\[ \mathbf{E} = -\nabla V = (-3, -4) \quad \text{(in units of V/m)} \]

### Step 2: Calculate the Force on the Particle
The force \( \mathbf{F} \) acting on the particle with charge \( Q = 1 \mu C = 1 \times 10^{-6} C \) is given by:
\[ \mathbf{F} = Q \mathbf{E} \]
Substituting the values:
\[ \mathbf{F} = (1 \times 10^{-6}) \cdot (-3, -4) = (-3 \times 10^{-6}, -4 \times 10^{-6}) \quad \text{(in N)} \]

### Step 3: Calculate the Acceleration
Using Newton's second law \( \mathbf{F} = m \mathbf{a} \), where \( m = 10 \, \text{kg} \):
\[ \mathbf{a} = \frac{\mathbf{F}}{m} = \left( \frac{-3 \times 10^{-6}}{10}, \frac{-4 \times 10^{-6}}{10} \right) = (-3 \times 10^{-7}, -4 \times 10^{-7}) \quad \text{(in m/s}^2\text{)} \]

### Step 4: Determine the Time to Cross the X-axis
The particle starts from rest at point \( (2, 3.2) \) m and crosses the x-axis when \( y = 0 \). The initial velocity \( u_y = 0 \) and the displacement in the y-direction \( s_y = 3.2 \) m. Using the equation of motion:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 \]
Substituting the values:
\[ 3.2 = 0 + \frac{1}{2} (-4 \times 10^{-7}) t^2 \]
Solving for \( t^2 \):
\[ 3.2 = -2 \times 10^{-7} t^2 \implies t^2 = \frac{3.2}{2 \times 10^{-7}} = 1.6 \times 10^7 \implies t = \sqrt{1.6 \times 10^7} \approx 4000 \, \text{s} \]

### Step 5: Calculate the Velocity Components When Crossing the X-axis
Using the equation for velocity:
\[ v_y = u_y + a_y t \]
Substituting the values:
\[ v_y = 0 + (-4 \times 10^{-7}) \cdot 4000 = -1.6 \times 10^{-3} \, \text{m/s} \]
For the x-component, since there is no acceleration in the x-direction:
\[ v_x = u_x + a_x t = 0 + (-3 \times 10^{-7}) \cdot 4000 = -1.2 \times 10^{-3} \, \text{m/s} \]

### Step 6: Calculate the Magnitude of the Total Velocity
The total velocity \( v \) when the particle crosses the x-axis is given by:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(-1.2 \times 10^{-3})^2 + (-1.6 \times 10^{-3})^2} = \sqrt{1.44 \times 10^{-6} + 2.56 \times 10^{-6}} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \, \text{m/s} \]

### Final Answer
The velocity of the particle when it crosses the x-axis is:
\[ \boxed{2 \times 10^{-3} \, \text{m/s}} \]