A particle is moving in a circular path of radius a under the action of an attractive potential `U=-(k)/(2r^(2))`. Its total energy is :

Options

A. `-(3)/(2)(k)/(a^(2))`
B. `-(k)/(4a^(2))`
C. `(k)/(2a^(2))`
D. Zero

Answer

To find the total energy of a particle moving in a circular path under the influence of a given potential, we can follow these steps:

### Step 1: Identify the potential energy
The potential energy \( U \) is given as:
\[ U = -\frac{k}{2r^2} \]
For a circular path of radius \( a \), we substitute \( r \) with \( a \):
\[ U = -\frac{k}{2a^2} \]

### Step 2: Calculate the force from the potential energy
The force \( F \) acting on the particle can be found using the relation:
\[ F = -\frac{dU}{dr} \]
Calculating the derivative:
\[ F = -\frac{d}{dr}\left(-\frac{k}{2r^2}\right) = \frac{k}{r^3} \]
Substituting \( r = a \):
\[ F = \frac{k}{a^3} \]

### Step 3: Relate the force to centripetal motion
For a particle moving in a circular path, the centripetal force is given by:
\[ F = \frac{mv^2}{a} \]
Setting the two expressions for force equal gives:
\[ \frac{mv^2}{a} = \frac{k}{a^3} \]
From this, we can solve for \( mv^2 \):
\[ mv^2 = \frac{k}{a^2} \]

### Step 4: Calculate the kinetic energy
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2}mv^2 \]
Substituting the expression for \( mv^2 \):
\[ K = \frac{1}{2}\left(\frac{k}{a^2}\right) = \frac{k}{2a^2} \]

### Step 5: Calculate the total energy
The total energy \( E \) of the particle is the sum of its kinetic energy and potential energy:
\[ E = K + U \]
Substituting the values we found:
\[ E = \frac{k}{2a^2} - \frac{k}{2a^2} = 0 \]

### Conclusion
Thus, the total energy of the particle is:
\[ \boxed{0} \]