A negative charge `Q` is distributed uniformly in volume of a sphere is radius R and a point charge particle (may be negative or positive) is present on the surface of this sphere then variation of escape velocity `(v_(s))` of charge 'q' as a function of 'q' will be [negect gravitational interaction].

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Answer

To solve the problem, we need to analyze the escape velocity of a charge \( q \) placed on the surface of a uniformly charged sphere with a negative charge \( Q \) distributed throughout its volume.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( v_s \) is the minimum velocity required for an object to escape the gravitational or electrostatic influence of another object. In this case, we are dealing with electrostatic forces since we are neglecting gravitational interactions.

2. **Energy Conservation Principle**:
The total mechanical energy at the point of escape must equal zero. The kinetic energy of the charge \( q \) when it is at the surface of the sphere is given by:
\[ KE = \frac{1}{2} m v_s^2 \]
where \( m \) is the mass of the charge \( q \) and \( v_s \) is the escape velocity.

3. **Electrostatic Potential Energy**:
The electrostatic potential energy \( U \) at the surface of the sphere due to the charge \( Q \) is given by:
\[ U = \frac{k Q q}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge distributed in the sphere, \( q \) is the charge at the surface, and \( R \) is the radius of the sphere.

4. **Setting Up the Equation**:
According to the conservation of energy:
\[ \frac{1}{2} m v_s^2 + \frac{k Q q}{R} = 0 \]
Rearranging gives:
\[ \frac{1}{2} m v_s^2 = -\frac{k Q q}{R} \]

5. **Solving for Escape Velocity**:
Now, we can express \( v_s \):
\[ v_s^2 = -\frac{2 k Q q}{m R} \]
Taking the square root:
\[ v_s = \sqrt{-\frac{2 k Q q}{m R}} \]

6. **Analyzing the Sign of \( Q \)**:
- If \( Q \) is negative, then \( -Q \) is positive, and since \( q \) can be either positive or negative, we need to analyze both cases:
- If \( q \) is positive, \( v_s \) will be real and positive.
- If \( q \) is negative, \( v_s \) will be zero because the product \( Qq \) will be positive, leading to a zero escape velocity.

7. **Conclusion**:
The escape velocity \( v_s \) is proportional to the square root of \( |q| \) when \( Q \) is negative and \( q \) is positive. Therefore, we can conclude that:
- For \( Q < 0 \) and \( q > 0 \), \( v_s \) is proportional to \( \sqrt{q} \).
- For \( Q < 0 \) and \( q < 0 \), \( v_s = 0 \).

### Final Result:
The variation of escape velocity \( v_s \) as a function of charge \( q \) is:
- \( v_s \propto \sqrt{|q|} \) for \( q > 0 \)
- \( v_s = 0 \) for \( q < 0 \)