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 [neglect gravitational interaction].

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Answer

To solve the problem of finding the variation of escape velocity \( v_s \) of a charge \( q \) as a function of \( q \) when a negative charge \( Q \) is uniformly distributed in a sphere of radius \( R \), we can follow these steps:

### Step 1: Understanding the System
We have a uniformly distributed negative charge \( Q \) within a sphere of radius \( R \). A point charge \( q \) is located on the surface of this sphere. We need to determine how the escape velocity \( v_s \) of the charge \( q \) varies with \( q \).

### Step 2: Use Energy Conservation
The principle of conservation of mechanical energy states that the total mechanical energy at the initial point (when the charge is at the surface) must equal the total mechanical energy at the point of escape (where the potential energy is zero).

The equation can be written as:
\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]

### Step 3: Write the Energy Equation
Initially, when the charge \( q \) is at the surface of the sphere, its kinetic energy is \( \frac{1}{2} m v_s^2 \) and the potential energy due to the charge \( Q \) is given by:
\[ U_i = -\frac{k Q q}{R} \]
where \( k \) is Coulomb's constant. At the point of escape, the kinetic energy is \( 0 \) and the potential energy is also \( 0 \).

Thus, we can write:
\[ \frac{1}{2} m v_s^2 - \frac{k Q q}{R} = 0 \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ \frac{1}{2} m v_s^2 = \frac{k Q q}{R} \]
Multiplying both sides by \( 2 \) leads to:
\[ m v_s^2 = \frac{2 k Q q}{R} \]

### Step 5: Solve for Escape Velocity \( v_s \)
Now, we can solve for the escape velocity \( v_s \):
\[ v_s^2 = \frac{2 k Q q}{m R} \]
Taking the square root gives:
\[ v_s = \sqrt{\frac{2 k Q q}{m R}} \]

### Step 6: Analyze the Relationship
From the equation \( v_s = \sqrt{\frac{2 k Q q}{m R}} \), we can see that \( v_s \) is proportional to the square root of \( q \):
\[ v_s \propto \sqrt{q} \]

### Conclusion
Thus, the variation of escape velocity \( v_s \) as a function of charge \( q \) is given by:
\[ v_s \propto \sqrt{q} \]