A thin plastic rod \(20 \mathrm{cm}\) long carries \(3.2 \mathrm{nC}\) distributed uniformly over its length. (a) If the rod is bent into a ring, find the potential at its center, (b) If it's bent into a semicircle, find the potential at the center (i.e., at the center of the circle of which the semicircle is part).

Linear charge density, represented by the Greek letter \( \lambda \), is a measure of the amount of electric charge per unit length along a line, such as a wire, rod, or edge. It is calculated by dividing the total charge \( Q \) by the length \( L \) over which the charge is distributed. Mathematically, the formula is represented as:
\[ \lambda = \frac{Q}{L} \]
Understanding linear charge density is essential because it affects the electric potential and field around a charged object. For example, if you have a plastic rod with a set amount of charge on it, and you know the length of the rod, you can calculate the linear charge density and subsequently use this information to find out how the charge affects the space around it. This becomes particularly important when the rod's shape is altered, as seen in the exercises where the rod is bent into different forms.

The electric potential at the center of a charged ring is an interesting case, as each point charge on the ring contributes to the potential at the center. Since all points on a ring are equidistant from the center, the ring's symmetry means that the potential due to each small segment of charge is the same. The total electric potential due to a ring can be found by integrating these small potentials from each segment over the entire circumference of the ring.
To simplify, if a ring has a uniform linear charge density, \( \lambda \), and we know the constant \( k \) represents Coulomb's constant \( (9 \times 10^9 \mathrm{Nm^2/C^2}) \), we can calculate the potential \( (V) \) at the center. We use the integral of the electric potential due to a point charge \( dV = \frac{k \lambda dl}{r} \), where \( r \) is the radius of the ring and \( dl \) is an infinitesimally small segment of the ring's circumference. Because \( r \) is constant for a ring, the integral simplifies to multiplying the circumference of the ring by \( k \lambda \), yielding a formula:
\[ V = 2\pi k \lambda \]

Calculating the electric potential at the center of a semicircle formed from a charged rod is somewhat similar to finding the potential at the center of a full ring, but with a key difference. The total potential is contributed by only half the circumference of the original circle. Therefore, instead of taking the full circumference into account, we consider only a half, which is \( \pi r \) for a semicircle.
Given a uniform linear charge density, \( \lambda \), we use the formula for the electric potential due to a point charge \( dV \) over the half circumference. The calculation is an integral of \( \frac{k \lambda dl}{r} \), but because we are only integrating over half the circumference, the potential at the center of a semicircle is:
\[ V = \pi k \lambda \]
Essentially, the potential at the center of a semicircle is half of what it would be at the center of a full circle (or ring) because you're integrating over only half the distance. It's a subtle but important distinction that can greatly affect the outcome when dealing with electrostatics.