Calculate the electric dipole moment of a system comprising of a charge + q distributed uniformly on a semicircular arc or radius R and a point charge – q kept at its centre.

Answer

To calculate the electric dipole moment of the given system, we will follow these steps:

### Step 1: Understand the System
We have a semicircular arc of radius \( R \) with a uniform positive charge \( +q \) distributed along it, and a point charge \( -q \) located at the center of the semicircle.

### Step 2: Define the Dipole Moment
The electric dipole moment \( \vec{p} \) is defined as:
\[ \vec{p} = \sum \vec{d} \cdot q \]
where \( \vec{d} \) is the displacement vector from the negative charge to the positive charge.

### Step 3: Set Up the Coordinate System
We will place the semicircular arc in the xy-plane. The center of the semicircle (where the negative charge \( -q \) is located) will be at the origin (0, 0). The positive charge is distributed along the semicircular arc.

### Step 4: Consider an Element of Charge
Take an infinitesimal element of charge \( dq \) on the semicircular arc. The length of the arc corresponding to an angle \( d\theta \) is given by:
\[ dq = \frac{q}{\pi R} \cdot R d\theta = \frac{q}{\pi} d\theta \]
This is because the total charge \( q \) is uniformly distributed over the semicircular arc of length \( \pi R \).

### Step 5: Calculate the Dipole Moment Contribution from \( dq \)
The position of the charge \( dq \) in polar coordinates can be expressed as:
\[ x = R \cos \theta, \quad y = R \sin \theta \]
The dipole moment contribution \( d\vec{p} \) from this charge \( dq \) can be expressed as:
\[ d\vec{p} = R \cdot dq \cdot \hat{r} \]
where \( \hat{r} \) is the unit vector pointing from the negative charge to the charge \( dq \). The direction of \( \hat{r} \) is along the line connecting the negative charge at the origin to the element \( dq \).

### Step 6: Resolve the Dipole Moment into Components
The dipole moment \( d\vec{p} \) can be resolved into its x and y components:
\[ d\vec{p}_x = d\vec{p} \cos \theta, \quad d\vec{p}_y = d\vec{p} \sin \theta \]
However, due to symmetry, the contributions to the y-component from opposite sides of the semicircle will cancel out, leading to:
\[ \text{Net } d\vec{p}_y = 0 \]

### Step 7: Integrate to Find the Total Dipole Moment
The total dipole moment will be the integral of the x-component of \( d\vec{p} \) over the semicircle:
\[ \vec{p} = \int_0^{\pi} d\vec{p}_x = \int_0^{\pi} R \cdot dq \cdot \cos \theta \]
Substituting \( dq \):
\[ \vec{p} = \int_0^{\pi} R \cdot \frac{q}{\pi} d\theta \cdot \cos \theta \]

### Step 8: Evaluate the Integral
The integral becomes:
\[ \vec{p} = \frac{Rq}{\pi} \int_0^{\pi} \cos \theta \, d\theta \]
The integral of \( \cos \theta \) from 0 to \( \pi \) is:
\[ \int_0^{\pi} \cos \theta \, d\theta = [\sin \theta]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \]
Thus,
\[ \vec{p} = 0 \]

### Conclusion
The total electric dipole moment of the system is:
\[ \vec{p} = 0 \]