Saved Bookmarks
| 1. |
Two charges -q and +q are located at points (0,0-a) and (0,0-a), respectively. (a) What is the electrostatic potential at the points (0,0,z) and (x,y,0) ? (b) Obtain the dependence of potential on the distance r of a point from the origin when r//a gt gt1. (c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis ? |
|
Answer» <P> Solution :Here - q is at (0,0 -a) and + q is at (0,0 a)(a) Potential at (0,0,z) would be `V = (1)/(4 pi epsilon_(0)) ((-q)/(z + a)) + (1)/(4 pi epsilon_(0)) ((q)/(z - a))` `= (q(z + a - z + a))/(4 pi epsilon_(0) (z^(2) - a^(2))) = (q^(2) a)/(4 pi epsilon_(0) (Z^(2) - a^(2)))` `V = (P)/(4 pi epsilon_(0) (z^(2) - a^(2)))` potential at (x,y,0) i.e., at a point 1 to z axis where charges are located is zero We have proved that `V = (P cos THETA)/(4 pi epsilon_(0) (r^(2) - a^(2) cos^(2) theta))` if `(r )/(a) gt gt 1` then `a lt lt r :. V = (P cos theta)/(4 pi epsilon_(0) r^(2))` `:. V = (1)/(r^(2))` i.e., potential is inversly PROPORTIONAL to square of the distance (c ) Potential at (5,0,0) is `V_(1) = (-q)/(4 pi epsilon_(0)) (1)/(sqrt((5 - 0)^(2)) + (-a)^(2)))` `+ (q)/(4 pi epsilon_(0)) (1)/(sqrt((5 - 0)^(2) + a^(2)))` `= (-q)/(4 pi espilon_(0)sqrt(25 + a^(2))) + (q)/(4 pi epsilon_(0) sqrt(25 + a^(2))) = 0` Potential at (-7, 0,0) `V_(2) = (-q)/(4 pi epsilon_(0)) (1)/(sqrt((-7 -0)^(2) + a^(2)))` = zero As work done = charge `(V_(2) - V_(1))` W = zero As work done by electrostatic field is independent of the path connecting the two points THEREFORE work done will CONTINUE to be zero along every path. |
|