Problem 68 - University Physics with Modern Physics

Question

A thin insulating rod is bent into a semicircular arc of radius \(a\), and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.

Step-by-Step Solution

Answer

The potential at the center is \(\frac{k_e Q}{a}\).

Step 1: Understand the Concept of Electric Potential

The electric potential at a point due to a charge distribution is the work done to bring a unit positive charge from infinity to that point. For a continuous charge distribution, this involves integrating over the charge distribution.

Step 2: Identify the Charge Element

The charge is uniformly distributed along the semicircular arc. Consider a small charge element, \(dq\), on the arc. The linear charge density, \(\lambda\), is given by \(\lambda = \frac{Q}{\pi a}\), where \(Q\) is the total charge and \(\pi a\) is the length of the semicircular arc.

Step 3: Setup the Expression for Potential

The potential \(dV\) at the center due to the charge element \(dq\) at a distance \(r = a\) (radius of the semicircle), is given by \(dV = \frac{k_e \cdot dq}{a}\) where \(k_e\) is Coulomb's constant.

Step 4: Substitute the Charge Element

Since \(dq = \lambda \, dl = \frac{Q}{\pi a} \, dl\) and the total potential \(V\) is the integral of \(dV\) over the semicircle, rewrite the expression as \(V = \int dV = \int_0^{\pi a} \frac{k_e \cdot dq}{a} = \int_0^a \frac{k_e \cdot \lambda \, dl}{a}\).

Step 5: Integrate Along the Semicircle

As \(\lambda = \frac{Q}{\pi a}\), substitute to get \(V = \frac{k_e \cdot Q}{\pi a} \int_0^{\pi a} \frac{dl}{a} \), since \dl\ is just along the arc of length \(\pi a\). This simplifies to \(V = \frac{k_e \cdot Q}{\pi a^2} \cdot \pi a = \frac{k_e \, Q}{a}\).

Step 6: Conclude with Result

Thus, the electric potential at the center of curvature of the semicircular arc due to the uniformly distributed charge is \(\frac{k_e \, Q}{a}\).

Key Concepts

Semicircular Arc

In physics, a semicircular arc refers to a half-circular shape which often serves as an essential component in problems involving circular symmetry. The semicircle has a defined radius, denoted typically by the letter \( a \), which describes the distance from the center of the full circle to any point on the arc. This property becomes significant when calculating electric potentials along the arc. When a charge distribution along a semicircular arc is considered, it is crucial to understand that every small segment of the arc contributes to the overall electric potential at a specific point, like the center of curvature in this case. The geometry of a semicircular arc facilitates the symmetrical distribution of effects due to its natural curve.

Uniform Charge Distribution

A uniform charge distribution occurs when a charge \( Q \) is spread evenly across a length, surface, or volume. In the case of a semicircular arc, since it is a one-dimensional problem, the charge distribution is linear. The total charge \( Q \) is uniformly dispersed along the length of the arc, which is \( \pi a \), being the half-circumference of a circle with radius \( a \).To describe this distribution mathematically, we use the linear charge density \( \lambda \), which is defined as the total charge divided by the length over which it is distributed:

• \( \lambda = \frac{Q}{\pi a} \)

This expression clarifies that each infinitesimally small length of the arc holds an equal amount of charge, leading to straightforward calculations of electric fields and potentials.

Continuous Charge Distribution

The concept of continuous charge distribution is pivotal when dealing with systems where charge is not confined to discrete points but is spread over a continuous path, line, area, or volume. In scenarios involving electric fields and potentials, this requires an integrative approach. Instead of summing discrete charges, calculus allows us to integrate over the charge distribution to find the total electric potential or field.With the given arc problem, the charge is not isolated but rather smeared uniformly across the semicircle. Thus, to compute the electric potential, we must take into account every infinitesimal charge element \( dq \) distributed along the arc. The potential at a given point, such as the center of curvature, hence involves:

• Breaking the arc into infinitesimally small elements each possessing a charge \( dq \)
• Summing the contributions from these infinitesimal charges by integration across the arc

This continuous approach captures the collective effect of the distributed charge, allowing for accurate calculation of the resultant electric potential.