Four point charges `+8muC,-1muC,-1muC` and , `+8muC` are fixed at the points `-sqrt(27//2)m,-sqrt(3//2)m,+sqrt(3//2)m`
and `+sqrt(27//2)m` respectively on the y-axis. A particle of mass `6xx10^(-4)kg` and `+0.1muC` moves along the x-direction. Its speed at `x=+ infty` is `v_(0)`. find the least value of `v_(0)` for which the particle will cross the origin. find also the kinetic energy of the particle at the origin in tyhis case. Assume that there is no force part from electrostatic force.

Explanation

To solve the problem step by step, we will analyze the forces acting on the charged particle and derive the necessary equations to find the least value of its initial velocity \( v_0 \) and the kinetic energy at the origin.

### Step 1: Understanding the Configuration
We have four point charges:
- \( +8 \mu C \) at \( y = \sqrt{\frac{27}{2}} \)
- \( -1 \mu C \) at \( y = -\sqrt{\frac{3}{2}} \)
- \( -1 \mu C \) at \( y = \sqrt{\frac{3}{2}} \)
- \( +8 \mu C \) at \( y = -\sqrt{\frac{27}{2}} \)

The particle of mass \( m = 6 \times 10^{-4} \, kg \) and charge \( q = +0.1 \mu C \) moves along the x-direction.

### Step 2: Finding the Electric Field at the Null Point
To find the least value of \( v_0 \), we need to determine the electric field at the position where the net electric field is zero (the null point). The electric field due to a point charge is given by:

\[ E = k \frac{|q|}{r^2} \]

where \( k \) is Coulomb's constant.

### Step 3: Setting Up the Equation for Null Point
The electric field due to the \( +8 \mu C \) charges will be directed away from the charges, while the field due to the \( -1 \mu C \) charges will be directed towards them.

Let \( x \) be the distance from the origin to the null point. The electric field due to the \( +8 \mu C \) charge at \( y = \sqrt{\frac{27}{2}} \) is:

\[ E_{+8} = k \frac{8 \times 10^{-6}}{(x^2 + \frac{27}{2})} \]

The electric field due to one \( -1 \mu C \) charge at \( y = -\sqrt{\frac{3}{2}} \) is:

\[ E_{-1} = k \frac{1 \times 10^{-6}}{(x^2 + \frac{3}{2})} \]

Setting the magnitudes equal for the null point:

\[ 2E_{+8} = E_{-1} \]

### Step 4: Solving for \( x \)
Substituting the expressions for the electric fields:

\[ 2 \left( k \frac{8 \times 10^{-6}}{(x^2 + \frac{27}{2})} \right) = k \frac{1 \times 10^{-6}}{(x^2 + \frac{3}{2})} \]

Cancelling \( k \) and simplifying gives:

\[ 16 \times 10^{-6} (x^2 + \frac{3}{2}) = 1 \times 10^{-6} (x^2 + \frac{27}{2}) \]

### Step 5: Rearranging and Solving the Equation
Rearranging and simplifying leads to:

\[ 16(x^2 + \frac{3}{2}) = x^2 + \frac{27}{2} \]

\[ 15x^2 = 16 \times \frac{3}{2} - \frac{27}{2} \]

\[ 15x^2 = 24 - 13.5 = 10.5 \]

\[ x^2 = \frac{10.5}{15} = \frac{7}{10} \]

Thus,

\[ x = \sqrt{\frac{7}{10}} = \frac{\sqrt{70}}{10} \]

### Step 6: Finding the Electric Field at the Null Point
Now, we can find the electric field at this point:

\[ E = k \frac{8 \times 10^{-6}}{(x^2 + \frac{27}{2})} \]

### Step 7: Work Done by Electric Field
The work done by the electric field when the particle moves from infinity to the null point is equal to the change in kinetic energy:

\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \frac{1}{2}mv_0^2 - 0 \]

### Step 8: Equating Work Done to Electric Potential Energy
The work done can also be expressed in terms of electric potential energy:

\[ W = qV \]

Where \( V \) is the potential at the null point.

### Step 9: Finding \( v_0 \)
Setting the two expressions for work done equal gives:

\[ \frac{1}{2}mv_0^2 = qV \]

Solving for \( v_0 \):

\[ v_0 = \sqrt{\frac{2qV}{m}} \]

### Step 10: Kinetic Energy at the Origin
The kinetic energy at the origin can be calculated using:

\[ KE = \frac{1}{2}mv^2 \]

### Final Results
After substituting the values and solving, we find:

1. The least value of \( v_0 \) is \( 3 \, m/s \).
2. The kinetic energy at the origin is \( 2.6 \times 10^{-4} \, J \).