Linear Charge Density - General Physics - Solved Past Paper, Exams of Physics

Download Linear Charge Density - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 2. (30 pts) An insulating rod carrying a uniformly linear charge density λ0 C/m is bent into the shape of part of a circle, as shown below. The thickness of the rod is exaggerated for clarity. The bottom end of the rod is at an angle θ0 below the negative x-axis and the upper end is at an angle θ0 above the negative x-axis. R dq θ dθ a) (5 pts) The angular displacement from one end of the indicated small piece rod (dq) to the other is dθ. How long is this piece of the rod? We need the length of the arc dθ. Arclength is angle multiplied by radius. Thus we have dℓ = R dθ b) (5 pts) Given your answer to (a) and the information provided above, what is the magnitude of the charge dq on the small piece of the rod? Charge on a length of rod would be charge density times length. Thus we have dq = λ0dℓ = λ0R dθ c) (5 pts) If one were to compute the electric field at the origin (located at the intersection of lines to the right of the rod), how far is the small piece of the rod from this location? The point is part of a circle, of which all points are a distance R from the origin. d) (5 pts) Now write down the magnitude of the electric field at the origin produced by this small piece of charge. dE = k dq r2 = kλ0R dθ R2 = kλ0dθ R e) (5 pts) Break this electric field into components if the indicated small piece of charge is at angular position θ as measured from the negative x-axis. Since the angle is given, we just note that the x component will be to the right and the y component will be down, giving: d ~E = kλ0dθ R cos θ ı̂ − kλ0dθ R sin θ ̂ f) (5 pts) Finally, write down the integral(s) you would perform to compute the net electric field at the origin. Just add the pieces that are on the rod, from −θ0 to +θ0: ~E = θ0∫ −θ0 kλ0dθ R cos θ ı̂ − θ0∫ −θ0 kλ0dθ R sin θ ̂