Question 5 please thank you Part 1: The Charged Semi-Circular HoopA thin plastic rod is bent into the shape of a semicircle of radius a. It is charged negativelywith -Q Coulombs of charge. Show how to derive the equation for the electric field at the originof the coordinate axes in the following diagram.81. Identify the position vector for a small portion of the source charge distribution and express itin plane polar coordinates using a unit vector in those coordinates.da2. 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 thesource to the field point.2=% circumferen4. Express dq in two ways. First in terms of a linear charge density λ and the integrationvariable. Then write it in terms of quantities given in the problem statement. PartialAnswer: dq = Ads where ds is a differential element of arc-length andnference of a half circle)5. Write the expression (the integral) for E' at the field point. Answer: E = k₁, d0f de6. Convert the previous expression back into Cartesian coordinates.πα

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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 dq 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 Tp =. 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 λ 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=%/cio (circumference of a half circle) 2 5. Write the expression (the integral) for E at the field point. Answer: E =k, Jax J = O ta fde