A particle of mass m and charge -q circulates around a fixed charge q in a circle radius under electrostatic force. The total energy of the system is (k= `(1/4piepsilon_0)`)

Options

A. `-(Kq^2)/(2r)`
B. `(Kq^2)/r`
C. `-(Kq^2)/r`
D. Zero

Answer

To solve the problem, we need to find the total energy of a system where a particle of mass \( m \) and charge \( -q \) is circulating around a fixed charge \( q \) in a circular path of radius \( r \) under the influence of electrostatic forces.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle**:
The particle with charge \( -q \) experiences an electrostatic force due to the fixed charge \( q \). This force provides the necessary centripetal force for the circular motion of the particle.

2. **Write the Expression for Electrostatic Force**:
The electrostatic force \( F \) between the two charges is given by Coulomb's law:
\[ F = \frac{k \cdot |q \cdot (-q)|}{r^2} = \frac{kq^2}{r^2} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

3. **Set Up the Centripetal Force Equation**:
The centripetal force required to keep the particle moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the speed of the particle.

4. **Equate the Electrostatic Force to the Centripetal Force**:
Since the electrostatic force provides the centripetal force, we can set them equal:
\[ \frac{kq^2}{r^2} = \frac{mv^2}{r} \]

5. **Solve for Velocity \( v \)**:
Rearranging the equation gives:
\[ mv^2 = \frac{kq^2}{r} \]
Thus,
\[ v^2 = \frac{kq^2}{mr} \]

6. **Calculate the Kinetic Energy \( KE \)**:
The kinetic energy of the particle is given by:
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{kq^2}{mr}\right) = \frac{kq^2}{2r} \]

7. **Calculate the Potential Energy \( PE \)**:
The potential energy \( U \) of the system due to the electrostatic interaction is given by:
\[ U = \frac{k \cdot q \cdot (-q)}{r} = -\frac{kq^2}{r} \]

8. **Find the Total Energy \( E \)**:
The total energy \( E \) of the system is the sum of the kinetic energy and potential energy:
\[ E = KE + PE = \frac{kq^2}{2r} - \frac{kq^2}{r} \]
Simplifying this gives:
\[ E = \frac{kq^2}{2r} - \frac{2kq^2}{2r} = -\frac{kq^2}{2r} \]

9. **Final Answer**:
Therefore, the total energy of the system is:
\[ E = -\frac{kq^2}{2r} \]

### Conclusion:
The total energy of the system is \(-\frac{kq^2}{2r}\), which corresponds to option 1 in the provided choices.