Saved Bookmarks
| 1. |
Two Charges -q each are separated by distance 2d. A third charge +q is kept mid point O. Findpotential energy of +q as a function of small distance x from O dueto q charges. Sketch P.E. v//s x and convince yourselfthat the charge at O is in an unstable equillibrium. |
|
Answer» Solution :In fig, two charges`-Q` each are shownat Aand B, where `AB = 2d`. A charge `+q` is kept at MID point O of `AB, OP = x` is small displacementof `+q` charge. Therefore, potential energy of `+q` charge at P due to the two charges `-q` each is `U = (1)/(4pi in_(0)) [-(q^(2))/(d + x) - (q_(2))/(d - x)] = (q^(2))/(4pi in_(0)) (2d)/(d^(2) - x^(2))`...(i) The potential energy (u) versus x GRAPH is as shown in Fig. Now `(dU)/(dx) = (-q^(2) (2d))/(4pi in_(0)) (2x)/(d^(2) - x^(2))`..(ii) At `x = 0`, From (i) `U_(0) = (-2q^(2))/(4pi in_(0) d)`, and from (ii), `(dU)/(dx) = 0` `:. x = 0` is an EQUILIBRIUM point Now `(d^(2) U)/(dx^(2)) = ((-2dq^(2))/(4pi in_(0))) [(2)/((d^(2) - x^(2))^(2) - (8x^(2))/(d^(2) - x^(2))^(3))]` `(d^(2) U)/(dx^(2)) = ((-2dq^(2))/(4pi in_(0))) (1)/((d^(2) - x^(2))^(3)) [2 (d^(2) - x^(2)) - 8x^(2)]` At `x = 0, (d^(2) U)/(dx^(2)) = ((-2dq^(2))/(4pi in_(0)))(1)/(d^(6)) (2d^(2)) = - (q^(2))/(pi in_(0) d^(3))` which is less thanzero. Hence, the equilibrium of chargeq at O is unstable.  
|
|