# Electric Field of a Charged Semicircle

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Lesson Transcript
Instructor: Matthew Bergstresser
Electric fields are generated by electrical charges. An extended object that is charged also generates an electric field. In this lesson, we'll derive the electric field at the radius of a charged semicircle.

## Electric Fields

Have you ever been to a concert with a laser light show or seen one on television? Light beams head out in all directions from their source. This is what an electric field might look like if you could see it. There are two types of electric charge, and they both generate electric fields. The field lines radiate out in all direction from positive charges and move in from all directions ending at negative charges.

## Ring Around the Origin

Let's start with several point charges arranged in a semicircle. Electric fields are vectors, which means they have a magnitude (a strength) and a direction. In this diagram, we can see the vector addition of all of the fields will leave only the y-components of the electric fields. All of the x-components of the fields cancel. This is illustrated in this diagram.

Now, let's pretend we add a lot more positive charges to our semicircle and turn it into a solid rod as seen in this diagram:

• dÎ¸ is a tiny change in angle. Think of it as a really small wedge of pumpkin pie.
• RdÎ¸ is a tiny segment of the circumference of the semicircle. Think of this as the tiny portion of the pumpkin pie crust from the wedge we just cut out of it. Multiplying the radius and the tiny angle dÎ¸ gives us a length.

Linear charge density, Î», is the charge-per-unit-length. It is useful because it is the same value for a tiny amount of the charged rod and the entire charged rod. The full length of the rod is half of the circumference of a circle with radius R giving us Ï€R. This allows us to write the expression for Î» as shown in Equation 1:

• Î» is linear charge density in coulombs-per-meter (C/m)
• Q is the total charge in coulombs (C)
• m is the length of the rod in meters (m)

Rearranging Equation 1, and solving for Q, we get Equation 2:

Since we're dealing with a tiny segment of the charged semicircle, Ï€R is replaced by RdÎ¸, and the charge associated with that tiny segment is given by Equation 3:

A tiny electric field is generated by the tiny charge dQ, and the equation for the magnitude (strength) of this electric field is given in Equation 4:

• dE is the tiny electric field in newtons-per-coulomb (N/C)
• k is a constant equal to 8.99 x 109 newton-meters-squared-per-square-coulomb (nm2 / C2)
• R is the radius of the distance from the tiny charge segment in meters (m)

Now, we can substitute the right side of Equation 3 into dQ of Equation 4 giving us Equation 5:

Just like when we were dealing with the individual charged particles and their x-components canceled, the same happens for the charged semicircle. All we need is the y-components of the electric fields shown in this diagram.

The y-component of the tiny electric field is given in Equation 6:

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