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प्रश्न
Two fixed, identical conducting plates (α and 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 (Figure). 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 and γ.
- Find the electric field acting on the plate γ before collision.
- Find the charges on β and γ after the collision.
- Find the velocity of the plate γ after the collision and at a distance d from the plate β.
उत्तर
a. The electric field at γ due to plate α is `- Q/(S2ε_0) hatx`
The electric field at γ due to plate β is `q/(S2ε_0) hatx`
Hence, the net electric field is E1 = `((Q - q))/(2ε_0S) (- hatx)`
b. During the collision plates β and γ are together and hence must be at one potential. Suppose the charge on β is q1 and on γ is q2. Consider a point O. The electric field here must be zero.
Electric field at 0 due to α = `- Q/(2ε_0S) hatx`
Electric field at 0 due to β = `q_1/(2ε_0S) hatx`
Electric field at 0 due to γ = `q_2/(2ε_0S) hatx`
∴ `(-(Q + q_2))/(2ε_0S) q_1/(2ε_0S)` = 0
⇒ q1 – q2 = Q
Further, q1 + q2 = Q + q
⇒ q1 = Q + q/2
And q2 = q/2
Thus the charge on β and γ are Q + q/2 and q/2, respectively.
c. Let the velocity be v at the distance d after the collision. If m is the mass of the plate γ, then the gain in K.E. over the round trip must be equal to the work done by the electric field. After the collision, the electric field at γ is
E2 = `- Q/(2ε_0S) hatx + ((Q + q/2))/(2ε_0S) hatx = (q/2)/(2ε_0S) hatx`
The work is done when the plate γ is released till the collision is F1d where F1 is the force on plate γ. The work done after the collision till it reaches d is F2d where F2 is the force on plate γ.
F1 = `E_1Q = ((Q - q)Q)/(2ε_0S)`
And F2 = `E_2 q/2 = (q/2)^2/(2ε_0S)`
∴ Total work done is `1/(2ε_0S) [(Q - q) Q + (q/2)^2]d = 1/(2ε_0S) (Q - q/2)^2d`
⇒ `(1/2)mv^2 = d/(2ε_0S) (Q - q/2)^2`
∴ `v = (Q - q/2)(d/(mε_0S))^(1/2)`
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