Question

A thin wire with positive charge evenly spread along its length is shaped into a semicircle. What is the direction of the electric field at the center of curvature of the semicircle? Explain.

Short answer

Short Answer: The net electric field at the center of curvature of a uniformly charged semicircle is directed towards the center of the semicircle (leftwards or in the negative x-direction).

Step-by-step solution

Understanding the Principle of Superposition

The principle of superposition states that the net electric field at any point due to multiple charges is equal to the vector sum of the electric fields due to each charge individually.

Breaking the Semircircle into Small Charge Elements

In order to apply the principle of superposition, we need to break the semicircle into infinitesimally small charge elements (dq) and then consider the electric field due to each of these small charge elements.

Calculating the Electric Field Due to a Charge Element

The electric field (dE) due to an infinitesimally small charge element (dq) located at angle theta on the semicircle can be written as dE = \frac{k dq}{r^2}, where k is the Coulomb's constant, r is the radius of the semicircle, and dq is the small charge element. To find the direction of the electric field, we will decompose the electric field into its horizontal (x-component) and vertical (y-component) at the center of curvature. dE_x = dE * cos(\theta) dE_y = dE * sin(\theta)

Integrating to Find the Total Electric Field

We will integrate the electric field components over the entire semicircle (from 0 to pi radians) to find the total electric field at the center of curvature. E_x = \int_0^\pi dE_x = k \int_0^\pi \frac{dq}{r^2} cos(\theta) E_y = \int_0^\pi dE_y = k \int_0^\pi \frac{dq}{r^2} sin(\theta) It's important to note that for any charge element located at angle theta, there will be another charge element at -theta with the same magnitude of electric field but opposite direction in the vertical (y) component. Therefore, when we integrate over the entire semicircle, the vertical component (E_y) will cancel out.

Analyzing the Total Electric Field

Since the vertical (y) component of the electric field disappears, only the horizontal component (E_x) remains. Therefore, the net electric field at the center of curvature of the semicircle is directed in the negative x-direction (leftwards), which means it is pointing towards the center of the semicircle.