Saved Bookmarks
| 1. |
A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin O. A negatively charged particle P is released from rest at the point `(0, 0, z_0)` where `z_0gt0`. Then the motion of P isA. periodic, for all values of `z_0` satisfying `0ltz_0ltoo`.B. simple harmonic, for all values of `z_0` satisfying `0ltz_0leR`C. approximately simple harmonic, provided `z_0lt lt R`D. such that P crosses O and continues to move along the negative z-axis toward `z = -oo`. |
|
Answer» let Q be the charge on the ring the negative `-q` is released from point `P(0,0,Z_(0))` the electric field at P due to the chaged ring will be along positive z-axis and its magnitude will e `E=(1)/(4piepsi)(QZ_(0))/((R^(2)+Z_(0)^(2)-4pt)^(3//2))` therefore force on charge P will be toward the centre shown, and its magnitude is `F_(e)=qE=(1)/(4piepsi_(0))(Qq)/((R^(2)+Z_(0)^(2))^(3//2))Z_(0)` ..(i) similarly when it crosses the origin the force is again toward center O. thus the motion of the particle is periodic for all values of `Z_(0)` lying between 0 and `infty` secondly if `Z_(0) lt lt R,(R^(2)+Z_(0)^(2))^(3//2)toR^(3)` `F_(e)=(1)/(4pepsi_(0))xx(Qq)/(R^(3))xxZ_(0)` [from eq (i)] that is the restoring force `F_(e)prop-Z_(0)`, thence the motion of the particle will e simple harmonic (here negative sign implies that the force is towards its means position). |
|