A particle having charge + q is fixed at a point O and a second particle of mass m and having charge `-q_(0)` moves with constant speed in a circle of radius r about the charge +q. the energy required to be supplied to the moving charge to increase radius of the path to 2r is `(q q_(0))/(n pi epsi_(0) r)`. Find the value of n.

Answer

To solve the problem, we need to find the energy required to increase the radius of the circular path of a charge from \( r \) to \( 2r \). We will follow these steps:

### Step 1: Calculate the initial potential energy (PE_initial)
The potential energy of the charge \( -q_0 \) at a distance \( r \) from the charge \( +q \) is given by:
\[ PE_{\text{initial}} = -\frac{q \cdot q_0}{4 \pi \epsilon_0 r} \]

### Step 2: Calculate the total energy (E_initial) at radius \( r \)
The total energy \( E_{\text{initial}} \) is the sum of the potential energy and the kinetic energy. The kinetic energy \( KE \) can be expressed in terms of the speed \( v_1 \):
\[ KE = \frac{1}{2} m v_1^2 \]
Using the centripetal force balance, we have:
\[ \frac{m v_1^2}{r} = \frac{k q q_0}{r^2} \]
Where \( k = \frac{1}{4 \pi \epsilon_0} \). Thus,
\[ v_1^2 = \frac{k q q_0}{m r} \]
Substituting this into the kinetic energy expression:
\[ KE = \frac{1}{2} m \left(\frac{k q q_0}{m r}\right) = \frac{k q q_0}{2 r} \]
Now, the total energy at radius \( r \) is:
\[ E_{\text{initial}} = PE_{\text{initial}} + KE = -\frac{q q_0}{4 \pi \epsilon_0 r} + \frac{k q q_0}{2 r} \]
Substituting \( k \):
\[ E_{\text{initial}} = -\frac{q q_0}{4 \pi \epsilon_0 r} + \frac{q q_0}{8 \pi \epsilon_0 r} = -\frac{q q_0}{8 \pi \epsilon_0 r} \]

### Step 3: Calculate the final potential energy (PE_final) at radius \( 2r \)
The potential energy at radius \( 2r \) is:
\[ PE_{\text{final}} = -\frac{q \cdot q_0}{4 \pi \epsilon_0 (2r)} = -\frac{q q_0}{8 \pi \epsilon_0 r} \]

### Step 4: Calculate the total energy (E_final) at radius \( 2r \)
Using the same method as before, we need to find the new speed \( v_2 \):
\[ \frac{m v_2^2}{2r} = \frac{k q q_0}{(2r)^2} = \frac{k q q_0}{4r^2} \]
Thus,
\[ v_2^2 = \frac{k q q_0}{m 2r} \]
Substituting this into the kinetic energy expression:
\[ KE_{\text{final}} = \frac{1}{2} m v_2^2 = \frac{1}{2} m \left(\frac{k q q_0}{m 2r}\right) = \frac{k q q_0}{4r} \]
Now, the total energy at radius \( 2r \) is:
\[ E_{\text{final}} = PE_{\text{final}} + KE_{\text{final}} = -\frac{q q_0}{8 \pi \epsilon_0 r} + \frac{q q_0}{4 \pi \epsilon_0 r} \]
This simplifies to:
\[ E_{\text{final}} = -\frac{q q_0}{8 \pi \epsilon_0 r} + \frac{2 q q_0}{8 \pi \epsilon_0 r} = \frac{q q_0}{8 \pi \epsilon_0 r} \]

### Step 5: Calculate the energy required (ΔE)
The energy required to increase the radius from \( r \) to \( 2r \) is:
\[ \Delta E = E_{\text{final}} - E_{\text{initial}} = \frac{q q_0}{8 \pi \epsilon_0 r} - \left(-\frac{q q_0}{8 \pi \epsilon_0 r}\right) = \frac{q q_0}{4 \pi \epsilon_0 r} \]

### Step 6: Relate to the given expression
The problem states that the energy required is given by:
\[ \Delta E = \frac{q q_0}{n \pi \epsilon_0 r} \]
Equating the two expressions:
\[ \frac{q q_0}{4 \pi \epsilon_0 r} = \frac{q q_0}{n \pi \epsilon_0 r} \]
From this, we can see that:
\[ n = 4 \]

### Final Answer
Thus, the value of \( n \) is \( 4 \).