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The class is about Electricity and Magnetism. The website wants exactly the perfect answer and in the same metric system units that are stated on

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The class is about Electricity and Magnetism. The website wants exactly the perfect answer and in the same metric system units that are stated on the calculators that are on the picture for the question, if not, it'll mark it wrong.

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The circular arc shown above has radius R = 12.2 cm and subtends a total angle 20 = 82, symmetric about the x axis. a. Suppose the arc carries a charge per unit length that varies with 0 as 1 = do cos 0, where @ is measured from the x axis. The value of do is 0.027 uC/m. Find the electric field at the origin. What is the magnitude IEl of the electric field? O. IEl = N/C abc sin(x Ox10 Submit Reset Reset All .Draw a sketch showing the infinitesimal electric field vector at the origin due to an infinitesimal element of arc of length R de. Use symmetry to figure out the direction of the electric field resulting from superposition of all the infinitesimal electric field vectors from each infinitesimal element of arc. .Use Coulomb's law to write an expression for the appropriate vector component of the infinitesimal electric field at the origin due to an element of charge on the arc. Then integrate to find the field due to the whole arc. .Remember to use radians for angular units in your computations. What is the direction of the electric field? Give the angle @ in degrees measured counterclockwise from the positive x axis. abc sin(x) Ox10 Submit Reset Reset Allb. Suppose the arc carries a charge per unit length that varies with 0 as 1 = 10 sin 0, where @ is measured from the x axis. The value of do is 0.027 uC/m. Find the electric field at the origin. What is the magnitude IEl of the electric field? 5 or IEl = N/C abc sIn(X Ox10 Submit Reset Reset All What is the direction of the electric field? Give the angle O in degrees measured counterclockwise from the positive x axis. 5 10 abc sin(x) 0x10 Submit Reset Reset Allde the change 7 20 sino X Ere - ve change Easing Easing Part by for 2 = 20 sing rupper half is positive and lower half is negative charge. so The final electric field will be in Ve y-ames. X- components commcell each other. direction of E Is 270 from X-any's dEy = dEsing = no sino. sing. Rdo RX day = 20 Sing ado R so total electric field E = sinto do R 270 sino do R Do = 210 R 1- cozo ) do 2 270 R SL 2 0 ) - [sinzo ] ] R L - * sinzoo = 0.44 X 10 8 / 0. 257 7 - 0.99 = 0. 048 X 10 N/2

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