Two charged thin spherical shells share same center at the origin. The inner she

Question

Two charged thin spherical shells share same center at the origin. The inner shell with radius r1 has total charge of -Q (Q > 0) evenly distributed on the surface. The outer shell has total charge of Q again evenly distributed on the surface with a radius of r2. Find the electric potential everywhere as a function of r. What is the potential difference between two shells (Delta V = V(r2) - V(r2))? A particle that has mass m and positive charge q (q > 0) is placed at r = r2 (right inside of the second shell) with zero initial velocity. The particle will be repelled by the outer shell toward the inner shell. Find the velocity of the particle when it reaches the inner shell. Two charged thin cylindrical shells share same axis. The inner shell has constant surface charge density sigma (sigma > 0) and radius r1. The outer shell has constant surface charge density -sigma and radius r2. What is the potential difference between two shells (Delta V = V(r2) - V(r1)? A particle that

Explanation / Answer

1) A two Spherical shells of inner radius r1 and outer r2 and their charges are -Q and +Q

              take a Gaussian surface

                  integral of E . da = Q/eps0

                              E 4 pi r2 = Q/eps0

                                   E = (1/4pi eps0 ) Q / r2

               a) Electric potential   V = integral of E dr

                                            V = -(1/4pi epso) Q / r

                  b) at r1 ,     V1= -(1/4pi eps0) (-Q / r1)

                        at r2,       V2 = -( 1/4pi eps0) Q / r2

                             dV = V2 - V1

                                  dV = -(1/4pi eps0) Q ( 1/r2 + 1/r1)

               c) a particle of mass m and charge q repelled by outer surface towards inner surface, velocity of particle at r1 with intial velocity zero.

                                          change in potential energy = change in Kinetic energy

                           dU = q dV= - (1/4pi eps0) q Q (1/r2 + 1/r1)

                      dU = -( 1/ 4pi eps0) qQ ( 1/r2 + 1/r1)

                        dE = 0 - 1/2 mv2 = -1/2 mv2

                                        1/2 mv2 = (1/4pi eps0) qQ (1/r2 +1/r1)

                                               v = sqrt ( 1 / 2m pi eps0 ) qQ ( 1/r2 +1/r1)

2) Two cylindrical shell of inner radius r1 with surface charge sigma S and outer radius r2 with surface charge -S

                        E = S /eps0

                           V = S r / eps0

              a) potential difference between two shells,

                     V1 = S r1 /eps0

                     V2 = -S r2 / eps0

                     dV = V2 - V1 = -S( r2 + r1) / eps0

                          dV = -S( r1 + r2) /eps0

           b) a paticle of mass m and charge q moves from r1 to r2 with inital velocity zero and final velocity v

                               Change in potential energy

                              dU = q dV = -qS (r1 + r2)/eps0

                             d E = 0 - 1/2 mv2

                      change in potential energy = change in kinetic energy

                              - 1/2 mv2 = - qS (r1 + r2 )/ eps0

                                    v = sqrt (2qS(r1 + r2)/ m eps0

                 

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.