Part 1: The Charged Semi-Circular Hoop A thin plastic rod is bent into the shape of a semicircle of radius a. It is charged negatively with -Q Coulombs of charge. Show how to derive the equation for the electric field at the origin of the coordinate axes in the following diagram. 8 $3 1. Identify the position vector for a small portion of the source charge distribution and express it in plane polar coordinates using a unit vector in those coordinates. 2. Identify the position vector for the field point and express your answer as 7, = . Recall, the field point is at the origin. 3. Determine an expression (in plane polar coordinates) for the vector R that points from the source to the field point. 4. Express dq in two ways. First in terms of a linear charge density and the integration variable. Then write it in terms of quantities given in the problem statement. Partial Answer: dq = Ads where ds is a differential element of arc-length and 2=%/ciscunferm umference of a half circle) Jux 5. Write the expression (the integral) for E' at the field point. Answer: E= k₁fd0 πα 6. Convert the previous expression back into Cartesian coordinates. 0-0

Part 1: The Charged Semi-Circular Hoop A thin plastic rod is bent into the shape of a semicircle of radius a. It is charged negatively with -Q Coulombs of charge. Show how to derive the equation for the electric field at the origin of the coordinate axes in the following diagram. 8 $3 1. Identify the position vector for a small portion of the source charge distribution and express it in plane polar coordinates using a unit vector in those coordinates. 2. Identify the position vector for the field point and express your answer as 7, = . Recall, the field point is at the origin. 3. Determine an expression (in plane polar coordinates) for the vector R that points from the source to the field point. 4. Express dq in two ways. First in terms of a linear charge density and the integration variable. Then write it in terms of quantities given in the problem statement. Partial Answer: dq = Ads where ds is a differential element of arc-length and 2=%/ciscunferm umference of a half circle) Jux 5. Write the expression (the integral) for E' at the field point. Answer: E= k₁fd0 πα 6. Convert the previous expression back into Cartesian coordinates. 0-0