# Semicircular rod and Electric Potential Question

• kevtimc25
In summary, a thin, flexible rod of length 10 cm and charge 91 nC is bent into a semicircle. The electrical potential at the center is found to be kQ/r. To make the potential zero at the center, a charge Q1 is needed at coordinates (x,y). The potential at a point due to a charge does not depend on the presence of other charges, and since the rod is an insulator, this problem is simplified.
kevtimc25

## Homework Statement

A thin, flexible rod of length L = 10 cm carries charge Q = 91 nC uniformly along its length. The rod is then bent into a semicircle, as shown in the figure. Show all work and circle answers. a) Find the electrical potential at the center.
b) Now we want to place a single point charge so that the electric potential is zero at the center. What are the coordinates of this charge’s positions (x,y) and what is the charge Q1 needed in Coulombs?

(I can't extract the figure, but its basically a semicircular rod of charge with a point where the center of the full circle would be.)

## Homework Equations

U= kQq / r, λ = Q/L, C=2πr

## The Attempt at a Solution

Part a.) We take (from 0-L)∫ dU = ∫(k/r)dQ => dQ = λdL, ∫(k/r)λdL => kλ/r * L ]from0-L
= (kλ/r) * L = kQ/r (same as U) , And we calculate r from 2L = 2πr.

But wouldn't the point charge in b) also be affected by the rod, and in that case we would have to vary 'r' as well to find that charge's elect. potential?

Much appreciated!

(a) looks like what I'd do.
(b) the potential at a point due to a charge does not depend on the presence of other charges... as your own equation shows.

Got it, thanks!

... note: it would matter if the rod were a conductor, then it becomes quite a hard problem.
Fortunately you are told that the charge density is a constant - so it must be an insulator.

## 1. What is a semicircular rod and how does it relate to electric potential?

A semicircular rod is a conductive material in the shape of a half-circle. It is often used in experiments to demonstrate how electric potential is affected by the shape and positioning of conductive materials. When a voltage is applied to the rod, the electric field inside the rod will be uniform and perpendicular to the surface, resulting in a constant electric potential along the length of the rod.

## 2. How is the electric potential calculated for a semicircular rod?

The electric potential of a semicircular rod can be calculated using the equation V = kQ/r, where V is the electric potential, k is Coulomb's constant, Q is the charge on the rod, and r is the distance from the center of the rod. This equation assumes that the rod is infinitely long and has a constant charge distribution.

## 3. What factors can affect the electric potential of a semicircular rod?

The electric potential of a semicircular rod can be affected by the charge on the rod, the distance from the center of the rod, and the shape and positioning of the rod. Additionally, the presence of other conductive materials nearby can also impact the electric potential of the rod.

## 4. How does the electric potential change as you move along the length of a semicircular rod?

As you move along the length of a semicircular rod, the electric potential will remain constant, assuming that the rod has a uniform charge distribution and there are no other nearby conductive materials. This is because the electric field inside the rod is perpendicular to the surface and will not change as you move along the length.

## 5. What are some potential real-world applications of semicircular rods and electric potential?

Semicircular rods and electric potential have various applications in physics and engineering, such as in capacitor design and electrostatic shielding. They can also be used in experiments to demonstrate the principles of electric potential and how it is affected by different factors. In addition, understanding electric potential is crucial in the design and operation of electronic devices and power systems.

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