[western50]

A total charge Q = -0.6 ?C is distributed uniformly over a quarter circle arc of radius a = 8.7 cm as shown.

Here is what I got :
The linear charge density ? Along the arc is -0.0000043905 C/m
Ex, the value of the x-component of the electric field at the origin (x,y) = (0,0) is 454189.6552 N/C
Ey, the value of the y-component of the electric field at the origin (x,y) = (0,0) is 454189.6552 N/C

But I don't know how to do the following two questions:
1. How does the magnitude of the electric field at the origin for the quarter-circle arc you have just calculated comnpare to the electric field at the origin produced by a point charge Q = -0.6 ?C located a distance a = 8.7 cm from the origin along a 45o line as shown in the figure?
A. The magnitude of the field from the point charge is less than that from the quarter-arc of charge.
B. The magnitude of the field from the point charge is equal to that from the quarter-arc of charge
C. The magnitude of the field from the point charge is greater than that from the quarter-arc of charge

2. What is the magnitude of the electric field at the origin produced by a semi-circular arc of charge = -1.2 ?C, twice the charge of the quarter-circle arc? As shown in the link below:

[jcaron2]

Q1: Just to make visualization a little easier, either tilt your head 45 degrees to the right or picture the arc shifted by 45 degrees to the left so that it spans between 45 and 135 degrees (instead of 0 to 90 degrees like it is now). The point charge Q in the middle will now be at the 90 degree mark. Without doing any math, you can see by inspection that the direction of the electric field at the origin will be straight up in both cases. In the case of the arc, the portion of the charge which is close to the middle will be contributing fully toward the upward-directed field. Out near the edges of the arc, however, the portions of the charge on the right and left will each contribute a somewhat diminished y-component (a factor sqrt(2)/2 right at the ends of the arc, for example). Their x-components, however, will cancel. Thus, the total contribution of that portion of the charge toward the total field at the origin is less than in the case of the point charge, where that same portion of the total charge is directing ALL of its field in the positive y direction. If the arc spanned 180 degrees, as it does in question 2, you can easily see that the charge portions way out on the ends of the arc would be contributing NOTHING to the y-component of the electric field (or to the x-component either, for that matter, since the two sides cancel each other out).

You could also, of course, simply do the math for both scenarios here. You already did the difficult version - that of the arc of charge where you have to use calculus, and you gave the answer at the beginning of your question. Calculating the field from a point charge, as you obviously know, is as simple as plugging the value of Q into E=kQ/r^2.

Q2: You can see from inspection that for every bit of charge to the right of the origin, there's a corresponding bit to the left. Thus, the x-component of the field will be cancelled at the origin. Meanwhile, the y-components of the electrical field for the left- and right-hand portions of the charge will add together. Thus the total field will have no x-component, and the y-component will be double what you calculated at the beginning of the problem (i.e. it will be about 908000 N/C). You could also do this with simple superposition: You already calculated the electric field components for the right-half 90-degree arc at the beginning of the problem. The corresponding electric field components from the left-half 90-degree arc would be exactly the same, but with the opposite sign for the x-component. To get the electric field for the total 180-degree arc, you could just add the contributions from each half (a.k.a. superposition), giving you 0 for the x-component and 908000 N/C for the y-component.

Make sense?