# A Thin Semicircular Rod Like The One In Problem 4

In a problem similar to problem 4, the total charge of a thin semicircular rod is Q. The test charge C is equidistant from all points along the rod. The force on the test charges is Fr. The total charge Q is equal to the sum of the charges at point C. If the point is C, then the electric field at that location is also Q.

Assume that a thin semicircular rod, a semi-circular rod, has a total charge q. Now, place the negative point charge -Q at a distance d from the positive point charge. Let us suppose the total charge q is placed in a uniform fashion across the upper and lower surfaces. In this case, a negative charge, d, is placed at a distance d from the negative point charge -Q.

If a thin semicircular rod has a charge q, then the net electric field is located along a line from the element to the charge minus cube. Since the small charge is positively charged, it attracts the positive point charge. Therefore, a negative point charge will attract a positive point charge. This charge will move to the center of the rod. Hence, the net electric field is generated along the line from the negative to the positive point charge.

Consider a thin semicircular rod with a total charge q. In this problem, the charge q is uniformly distributed along the lower half. The negative point charge -Q is placed at the end of the negative point charge. When the negative point is attracted to the positive, the electric field will be directed along the line. A tangent line is also formed.

In a similar problem, the total charge of a thin semicircular rod is q. The negative point charge is placed on point C. The positive and negative points are opposite each other. Hence, the net electric field points along the line connecting the element and the charge minus cube. In a situation such as this, the q charge is placed at the same distance as point C.

A thin semicircular rod is a solid that has a total charge q. The negative point charge is placed on point C. The net electric field points along the line. The small charge will be attracted by the positive point charge. The test and the negative is the same in both cases. It is important to understand that the two points are opposite to each other. In this case, the q and a charge of a thick rod will be equidistant.

Now, imagine that a thin semicircular rod with a total charge q is bent into a semicircular rod. The negative point charge -Q is placed on the top of the rod. If the two points are parallel, then the negative point charge will be attracted to the other. So, a negative point charge is located on the top of the rod. It is equidistant from the -Q.

In this problem, a negative point charge is placed on a thin semicircular rod. The negative point charge will be attracted to the positive. Thus, the two points are in opposite directions. The negative part is on top of the -Q. The other half is on the left side of the negative point. The negative point charge is at the top of the rod. The two points are equidistant to each other.

The white component of the stress distribution on the test is on the charge C of the little charge. The net electric field points along the line between the element and the charge minus cube. The small negative point charges will be attracted to the positive point charges. If the negative point charge is on the opposite side of the test, the other will be attracted to it. This is the same for the other three points.