Question

4. A flexible plastic rod can be charged and bent into a semicircle. Using the method of "breaking the object" into many point charges and then integrating the electric field from those charges, derive an equation for the electric field components at the center of the semicircle for a rod of length L, bent into a semicircle of radius R, with charge Q. Hints: Use the angle for your position of each "point charge". The length of a small segment of the circle is ds = Rdo. Note that the distance from each "point charge to the point we are calculating the field is the same, specifically R. Can you quickly explain the y component of the electric field or do you need to make a calculation? Use your equation to find the electric field at the center of a semicircular charge of radius 0.20 cm and total charge - 25.0 nC.

Answer 1

Son charge a radius=R total length. I = 21TR - TR charse per un's rength. M -910) Take an elemery as an angle o from y axis. Angle subsettes be do n dex elemey on the centre - do de length of element in Rdo Rey charge on eurers, dzirdo te teret Electric field due to charge dgas centre dekade der denna Er & ALERNO - KARdo fino R2 - KAlanodo Ex = KA Ecoso = 2k1] R

dEy decoro u kadrdo coso R2 KAR coro do R2 КА Ғул Ел. соло во - ker scorodo Ey & Ich Einino Enes - Ex 24 = 2 kile 2 ko "TR.R 2 ko TRE Q = -25he = -2510-9 R 50.20cm -0.20 4182 m. T + (0-241022 3580.9 410 4 = 3.5814107 N(C)