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

Download Uniform Charge Density - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 2. (30 pts) An insulating rod carrying a uniform linear charge density -25 µC/m is bent into the shape of part of a circle with a radius of 2 m, as shown below. The thickness of the rod is exaggerated for clarity. The bottom end of the rod is at an angle 30◦ below the negative x-axis and the upper end is at an angle 30◦ above the negative x-axis. 30o 30o 2 m x y -25 µC/m a) (10 pts) What is the total charge on this rod? There are a couple ways to do this. The easiest, since the charge density is uniform, is to multiply by the length of the rod. A 60◦ arc of radius 2 m has a length of ℓ = (π/3)(2) = 2.09 m. The charge is thus Q = (−25 × 10−6)(2.09) = −52.4 × 10−6 C. You could also integrate (add up) all the little pieces of charge along the rod. This goes like Q = π/6 ∫ −π/6 λRdθ = λR(π/3) = (−25 × 10−6)(2)(π/3) = −52.4× 10−6 C Of course, you need to make sure that you use radians for the angles! b) (10 pts) Show that the potential at the origin is keQ/R where Q is the total charge on the rod and R is the radius of curvature. (You don’t need to use numbers here, and remember that it is not a point charge!) The key here is that the curved wire is not a point charge. Thus you cannot simply write down the an- swer. You need to show that you get the answer when you apply the correct physics and mathematics. Begin by finding the contribution to the electric potential from a small piece, dq: dV = k dq r We know that r = R here, and then that dq = λdℓ = λRdθ. So we have dV = kλRdθ R Then integrate over the appropriate angles V = θ0 ∫ −θ0 kλRdθ R = kλ(2θ0) Making the substitution that λ = Q/(2θ0R), we then obtain the answer, V = k ( Q 2θ0R ) (2θ0) = kQ R You could also note, in the integral above, that k and R are constants, giving V = k R ∫ dq = kQ R