Question

a great distance? FIGURE 23-29 Problem 34 +3Q 35. (II) A flat ring of inner radius Ri and outer radius R2 Fig. 23-30, carries a uniform surface charge density σ. Determine the elec- tric potential at points along the axis (the x axis). [Hint: Try substituting variables.] af. Try 0 FIGURE 23-30 Problem 35. 36. (II) A total charge Q is uniformly distributed on a thread of length . The thread forms a semicircle. What is the potential at the center? (Assume V - 0 at large distances.) 37. (II) A 12.0-cm-radius thin ring carries a uniformly distributed 15 0 uC charge. A small 75 snbere witl

Please solve number 35 and explain how to set up the problem. Thanks!

Answer 1

Given That let us consider a flat ring of inner radius is = R_T outer radius R_2 and the surface charge density of the ring is = sigma radius is = R small element of radius of ring is dR The charge distribution of small element is dQ = sigma (2 pi R) dR The potential due to small element ring at distance x on the x - axis dv = dQ/4 pi epsilon_0 squareroot x^2 + R^2 \r\n where dQ = sigma (2 pi R) dR) dv = sigma (2 pi R)dR/4 pi epsilon_0 squareroot x^2 + R^2 dv = sigma (R dR)/2 epsilon_0 squareroot x^2 + R^2 rightarrow 1 potential due to total ring at a distance x on the x-axis integrate the final equation dv limits R_1 = R_1 + 0 R_2 V = integral^R_2_R_1 dv rightarrow 2 Substitute dv in the above eqn 2 v = integral^R_2_R_1 sigma (R dR)/2 epsilon_0 squareroot x^2 + R^2 v = sigma/2 epsilon_0