Can someone please help me summarize and explain teh application part like this example for my ow

My problem I need an expiation and application and summary please. I solved it and got the work done down below

Example:

Module 22-6 A Point Charge in an Electric Field v Joe Gibb posted Jun 7, 2023 9:09 PM 3'): Subscribe Summary: ) > An electrostatic force acts on the particle, as given by F = qE. If charge q is positive, the force vector is in the same direction as the field vector. lf charge q is negative, the force vector is in the opposite direction (the minus sign in the equation reverses the force vector from the field vector). Question: An electron is accelerated eastward at 1.80 x 109 m/s2 by an electric field. Determine the field (a) magnitude and (bl direction. Solution: Acceleration of electron, a = 1. 80 X 109m/32 eastward 5': (1.80 X 109m/s2)s Mass of electron = 9. 11 x 1031169 Electron Charge = 1. 60 X 10190 > _) } > F = mg, and also,F = qE So, the electric field; 9.11x 1031kg) (1.80x109m/32)? 1.60x 10190 = (0. 0102N/0) E = 0. 0102N/C westward a) Magnitude of the electric field = 0.0102N/C b) Direction of electric field= Westward Application: Alternating electric field therapy is a type of electromagnetic therapy that uses low-intensity electric fields, between 100-300 kHz, which can disrupt cancer cell division and slow the growth of brain tumors. There is currently a US and Europe approved device for the treatment of recurrent glioblastoma. Reference: Mittal, S., Klinger, N. V., Michelhaugh, S. K., Barger, G. R., Pannullo, S. C., & Juhasz, C. (2018]. Alternating electric tumor treating fields for treatment of glioblastoma: rationale, preclinical, and clinical studies. Journal of neurosurgery, 128(2), 414421. https://doi.orgI10.3171/2016.9.JN516452 A less Step 1: Writing the given values Given, 9 = 0.575 UC = 0.575 x10-C L = = 30.3 (m = 0.303 m Step 2: Solving the problem The electric field at the centure of semicircle is, En = - RK q 1 L 2 > En = - 2x (9x10 9 ) X (0. 575 x 20-6 ) XT (0. 303 ) 2 > Ex = - 3. 54 x 109 N/C Hence the magnitude of the electric field is 3.54 x10" N/ e Answer : Option IV ) 3.59 * 20 9 N / C___.___ __ _____ _________ ________ r-- ___ _ ___..______ __ __ __ Firstly, before you solve the problem, explain in your own words using a few sentences the underlying physics concept(s) that applies to the problem you selected. Secondly, provide the problem-solving steps with some explanation of the steps. Cursory comments are fine for each step. The equation editor within the discussion question window is provided to properly display equations. Please use this rather than copying and pasting equations from another source. Thirdly, based on your underlying physics concept(s) think of or find another situation/example where those concepts come into play and describe this for your classmates. 0 You must use at minimum one source, with a proper APA citation, which cannot be your text book. 0 Be prepared to facilitate the discussion in terms of helping you and your classmates to better understand and appreciate the concept. Supply hyperlinks and citations for all outside sources used. A thin, uniformly charged rod 30.3 cm long is bent into a semicircle. The total charge on the rod is 0.575 ,u C. What is the magnitude of the electric eld at the center of the semicircle? 4.25 x 104m: 11.1 x 104N/c 5.31 x 104N/c 3.54 x 104 N/C 7.08 x 104m