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Two fixed, identical conducting plates (alpha and beta), each of surface area S are charged to -Q and q, respectively, where Q gt q gt 0. A third identical plate (gamma), free to move is located on the other side of the plate with charge Q at a distance d (figure). The third plate is released and collides with the plate beta. Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst beta and gamma. (a) Find the electric field acting on the plate gamma before collision. (b) Find the charges on beta and gamma after the collision. (c) Find the velocity of the plate gamma after the collision and at a distance d from the plate beta. |
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Answer» SOLUTION :(a). Find electric field at plate `gamma` before collision is equal to the sum of electric field at plate `gamma` due to plate `ALPHA and beta`. The electric field at plate `gamma` due to plate `alpha` is `E_(1)=(-Q)/(S(2epsi_(0)))` to the left. The electric field at plate `gamma` due to plate `beta` is `E_(2)=(q)/(S(2epsi_(0)))`, to the right HENCE, the ent electric field at plate `gamma` before collision, `E=E_(1)+E_(2)=(q-Q)/(S(2epsi_(0)))`, to the left if `Q gt q` (b) During collision, plates `beta and gamma` are together. Their potentials become same. Suppose charge on plate `beta ` is `q_(1)` and CAHRGE on plate `gamma` is `q_(2)`. At any point O, in between the two plates, the electric field must be zero. Electric field at O due to plate `alpha=(-Q)/(S(2epsi_(0)))`, to the left Electric field at O due to plate `beta=(q_(1))/(S(2epsi_(0)))`, to the right ltBrgt Electric field at O due to plate `gamma=(q_(2))/(S(2epsi_(0)))`, to the left As the electric field at O is zero, therefore ltBrgt `(Q+q_(2))/(S(2epsi_(0)))=(q_(1))/(S(2epsi_(0)))` `thereforeQ+q_(2)=q_(1)` `Q=q_1+q_2` . . .(i) As there is no loss of charge o collision, ltBrgt `Q+q=q_(1)+q_(2)` On solving eqs. (i) and (ii), we get `q_(1)=(Q+q//2)`=charge on plate `beta` `q_(2)=(q//2)`=charge on plate `gamma` (c) After collision, at adistance d from plate `beta`. Let the VELOCITY of plate `gamma` be v. After the collision, electric field at plate `gamma` is `E_(2)=(-Q)/(2epsi_(0)S)+((Q+q//2))/(2epsi_(0)S)=(q//2)/(2epsi_(0)S)` to the right just before collision, electric field at plate `gamma` is `E_(1)=(Q-q)/(2epsi_(0)S)` If `F_(1)` is force on plate `gamma` before collision, then `F_(1)=E_(1)Q=((Q-q)Q)/(2epsi_(0)S)` Total work done by the electric field is round trip movement of plate`gamma` `W=(F_(1)+F_(2))d` `=([(Q-q)Q+(q//2)^(2)]d)/(2epsi_(0)S)=((Q-q//2)^(2)d)/(2epsi_(0)S)` ltBrgt If m is mass of plate `gamma`, the KE gained by plate `gamma=(1)/(2)mv^(2)` According to work-energy principle, `(1)/(2)mv^(2)=W=((Q-q//2)^(2)d)/(2epsi_(0)S)` `gamma=(Q-q//2)((d)/(mepsi_(0)S))^(1//2)`. |
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