An electric dipole is kept on the axis of a uniformly charged ring at distance from the centre of the ring. The direction of the dipole moment is along the axis. The dipole moment is `p`, charge of the ring is `Q` & radius of the ring is `R`. The force on the dipole is

Options

A. `(4kPQ)/(3sqrt(3)R^(2))`
B. `(4kPQ)/(3sqrt(3)R^(3))`
C. `(2kPQ)/(3sqrt(3)R^(3))`
D. zero

Answer

To find the force on an electric dipole placed on the axis of a uniformly charged ring, we can follow these steps:

### Step 1: Understand the Configuration
We have a uniformly charged ring with total charge \( Q \) and radius \( R \). An electric dipole with dipole moment \( \mathbf{p} \) is placed along the axis of the ring at a distance \( d \) from its center. The dipole consists of two equal and opposite charges, \( +q \) and \( -q \), separated by a small distance \( 2a \).

### Step 2: Electric Field Due to the Ring
The electric field \( \mathbf{E} \) at a point along the axis of the ring (at distance \( d \) from the center) can be calculated using the formula:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Qd}{(d^2 + R^2)^{3/2}} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Force on the Dipole
The force \( \mathbf{F} \) on a dipole in an electric field is given by:
\[ \mathbf{F} = \mathbf{p} \cdot \nabla \mathbf{E} \]
However, since the dipole is placed at a distance much larger than its length, the force on the dipole can be approximated by considering the forces acting on the individual charges \( +q \) and \( -q \).

### Step 4: Analyzing Forces on Charges
The charge \( +q \) experiences a force \( \mathbf{F}_{+} = +q \mathbf{E} \) in the direction of the electric field, while the charge \( -q \) experiences a force \( \mathbf{F}_{-} = -q \mathbf{E} \) in the opposite direction. Therefore, the net force on the dipole is:
\[ \mathbf{F}_{net} = \mathbf{F}_{+} + \mathbf{F}_{-} = +q \mathbf{E} - q \mathbf{E} = 0 \]

### Step 5: Conclusion
Since the forces acting on the two charges of the dipole cancel each other out, the net force on the dipole is zero:
\[ \mathbf{F}_{net} = 0 \]

### Final Answer
The force on the dipole is \( \mathbf{F} = 0 \).

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