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An insulated small sphere carrying a positive charge is accelerated through a potential $V_{1}$ , producing a horizontal beam along the x-axis. The beam enters a perfectly uniform electric field parallel to the y-axis, produced by a parallel plate capacitor with potential difference, $V_{2}$ between the plates, which have separation d and are of length, L as shown in figured. Take $V_{1} = 2 V$ and $V_{2} = 1 V .$ Assume that collision of particle with the plates is perfectly elastic,

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If L = 100 d, the angle made by velocity of particle with the x-axis just after emerging from the capacitor is zero.

If L = 100 d, the number of collisions made by the particle with the plates of the capacitor is 25.

If L = 105d, the angle made by velocity of particle with the x-axis just after emerging from the capacitor is $\tan^{-} (\frac{1}{2})$ .

Irrespective of the value of L, the number of collisions between the particle and the upper plate of the capacitor is same.

answer is A, B, D.

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Detailed Solution

1/2 m v_x^2 = 2d

u_x = √(4Q/m) v_y = 1V

m.o. → u_x
↓
x-y

a_y = QE/m = Q/md , a_x = 0

d = 0 + 1/2 (Q/dm) t^2
= √(d^2 m / Q) = t

S_x = 4xt = √(4Q/m) × √(d^2 m / Q)
= 2d & V_y = Q/dm × √(d^2 m / Q)
V_y = Q/m

Number of collisions = L / 4d

t = d/u_x , V_y = (Q/dm) × (d/u_x), V_y = Q/mu_x

tan θ = Q / mV_x^2 = Q × m / 4Q
tan θ = 1/4

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