Two fixed, identical conducting plates (α &β ) , each of surface area S are charged to –Q and q, respectively, where Q > q > 0. A third identical plate (γ ), free to move is located on the other side of the plate with charge q at a distance d . The third plate is released and collides with the plate β . Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst β &γ .  (a) Find the electric field acting on the plate γ before collision.   (b) Find the charges on β and γ after the collision. 

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7 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- net electric field at plate γ due to two other plate

From plate 1 ,E1= - Q S 2 ε 0 t o t h e l e f t

From plate 2 ,E2= q S 2 ε 0 t o t h e r i g h t

Total electric field E= E1 + E2 = q - Q S 2 ε 0 to the left , if Q>q

electric field at o due to plate α = - Q S 2 ε 0 t o t h e l e f t

electric field at o due to plate β

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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