Saved Bookmarks
| 1. |
Two fixed, identical conductingplates (alpha and beta), each of surface area S are charged to -Q and q, respectively, where Q gt q gt 0. A thirdindentical plate (gamma), free to move is locatedon the other side of the plate with charge q at a distanced, fig. The third plate is released andcollidies with theplate beta. Assumethe collsion is electricand the time of collision is sufficient to redistributechargeamongst beta and gamma. (a) Find the electricfieldacting on the plate gamma before collision. (b) Find the charge on beta and gamma after the collision. (c) Find the velocity of the plate gammaafter the collision and at a distance d from the plate beta. |
|
Answer» Solution :The given set up is shown in FIG. (a) The electric field at plate `gamma` due to plate `alpha` is `E_(1) = - (Q)/(S (2in_(0))` , to the left The electricfield at plate `gamma` to plate `beta` is `E_(2) = (q)/(S(2 in_(0))` , to the right. HENCE, the net electric field at plate `lamma` beforecollision is `E = E_(1) + E_(2) = (q-Q)/(S (2in_(0))` , to the left if `Q gt q` (b) During collision, plates`beta and gamma` are together. Their potentials become same. Suppose chargeon plane`beta is q_(1)` and chargeon plate `gamma is q_(2)`. At any point 0, inbetween the two plates, Fig, the electric field must be zero. Electric field at 0 due to plate `alpha = ((-) Q)/(S (2in_(0)))` , to the left Electric field at 0 due to plate `beta = (q_(1))/(S(2in_(0)))` , to the right Electric field at 0 due to plate `gamma = (q_(2))/(S(2in_(0)))` , to the left As the electric field at 0 is zero, therefore , `(Q+q_(2))/(S(2in_(0))) = (q_(1))/(S(2 in_(0)))` `:.Q + q_(2) = q_(1) or Q = q_(1) - q_(2)`...(i) As there is no loss of charge on collision, `Q + q = q_(1) + q_(2)`...(ii) On solving (i) and (ii),we get`q_(1) = (Q + q//2)` = charge on plate `gamma`. (c) After collision,at a distance d from plate `beta`, let the velocity of plate`gamma` be v. After the collision, electric field at plate `gamma` is`E_(2) = (-Q)/(2 in_(0) S) + ((Q + q//2))/(2 in_(0) S) = (q//2)/(2 in_(0) S)` to the right. Just before collision, electric field at plate `gamma` is `E_(1) = (Q-q)/(2 in_(0) S)` If `F_(1)` is force on plate`gamma` before collision, then`F_(1) = E_(1) Q = ((Q-q)Q)/(2 in_(0) S)` Similarly, force `F_(2)` on plate `gamma` after collision,`F_(2) = E_(2) (q)/(2) = ((q//2)^(2))/(2in_(0) S)` Total work DONE by the electricfield in round trip movement of plate `gamma` `W = (F_(1) + F_(2))d = ([(Q-q) Q+ (q//2)^(2)] d)/(2in_(0) S) = ((Q-q//2)^(2) d)/(2in_(0) S)` If m is MASS of plate `gamma`, the KE gainedby plate `gamma = (1)/(2) mv^(2)` ACCORDINGTO work energy principle, `(1)/(2) mv^(2) = W= ((Q-q//2)^(2) d)/(2in_(0) S)` `v = (Q - q//2) ((d)/(m in_(0) s))^(1//2)` 
|
|