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 find the total energy of a system where a particle of mass \( m \) and charge \( -q \) circulates around a fixed charge \( q \) in a circular path of radius \( r \), we will calculate both the kinetic energy and potential energy of the system and then sum them up.

### Step-by-Step Solution:

1. **Identify the Forces:**
The particle with charge \( -q \) is in circular motion around the fixed charge \( q \). The force providing the centripetal acceleration is the electrostatic force between the two charges.

2. **Centripetal Force:**
The centripetal force \( F_c \) required to keep the particle in circular motion is given by:
\[ F_c = \frac{m u^2}{r} \]
where \( u \) is the linear velocity of the particle.

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

4. **Equating Forces:**
Since the electrostatic force provides the necessary centripetal force, we can set them equal:
\[ \frac{m u^2}{r} = \frac{k q^2}{r^2} \]

5. **Solving for Velocity:**
Rearranging the equation to solve for \( u^2 \):
\[ m u^2 = \frac{k q^2}{r} \implies u^2 = \frac{k q^2}{m r} \]

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

7. **Potential Energy:**
The potential energy \( PE \) of the system is given by:
\[ PE = \frac{k q (-q)}{r} = -\frac{k q^2}{r} \]

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

9. **Final Result:**
The total energy of the system is:
\[ E = -\frac{k q^2}{2r} \]

### Conclusion:
The total energy of the system is \( -\frac{k q^2}{2r} \).