Question

Circumference (0) To find the potential (voltage) at point (P) distant x (along the ring's axis) from the ring center, we can first find the E-field as we did in chapter 21, by considering the contributions to the E-field from an elemental charge on the ring. of the E-field that points in the direction of the Symmetry then showed that only the axis of the ring (at points on the axis) survives is easy to perform the integral-Eai) going along the ring axis from o to x, to get V (x). potential everywhere of a point charge. So for an Alternatively, we already know the elemental charge dq, the contribution to the potential at point P is: (where v+a Integrate this elemental contribution over all the charees in the ring, and obtain an expression for the potential (voltage) V at x (on the ring axis) (Hint: dq-pds where ρ is charge density, ds is elemental length: Also, x, a, r, do not vary as we perform the integration over the charges on the ring. Also, k-9x10 (SI units)) V(x) (ii) What is the voltage (the potential energy of +1C) at x -0.3 m, and at the center of the ring. voltage (at x. 0.3m) = V). Voltage at center of ring - (iv) What is the potential energy of a small ball(treated as a point charge) of mass 0.0015 kg and charge-3 x 10°C, located at x = 0.3 m. If the location of the ball is changed to the center of the ring, what is the new potential energy. Potential energy (at x -0.3 m)- J); Potential energy at center of ring- (v) If the small ballis released from rest at x -0.3 m, what is the speed of the ball at the center of the ring, assuming the ball (Hint: employ the mechanical energy conservation law: Ki+Ui-K+ U) is constrained to move only along the axis of the ring Velocity (m/s)

Answer 1

R = squarerootx^2 + a^2 a = radius of ring q = change on ring v(alpha) = k dq/squarerootx^2 + a^2 now, dq = q/2 pi a dl: dl is element along the ring v(x) = k/squarerootx^2 + a^2 integral^2 pi a_0 dl q/2 pi a v(x) = k q/squarerootx^2 + a^2 = 9 times 10^9 q/squarerootx^2 + a^2 (ii) potential energy of charge (q_1 = IC) at x = 0.3 cm and at center since, potential energy (P E) is = q_1 v(x) since, potential energy (P.E) is = q_1 v(x) therefore voltage (at x = 0.3) = Ic times 9 times 10^9 times q/squareroot(0.3)^2 + a^2 v voltage (at x = 0) or center of ring = IC times 9 times 10^9 times q/a v iv) q_1 = -3 times 10^-6 c at x = 0.3 and at center of POE (at x = 0.3) = -3 times 10^-6 times 9 times 10^9 times 2/squareroot 0.3^2 + a^2