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Distance between plates (s) = 5 cm = 5 × 10^{-2}m

Charge of ∝ particle (q) = 3.2 × 10^{-19}c

Mass of ∝ particle (m) = 6.68 × 10^{-27}kg

Time taken (t) = 2 × 10^{-6}sec

Electric field between the plates (E) = ?

We have

s = ut + \(\frac{1}{2} \)at^{2}

s = \(\frac{1}{2} \)at^{2} (∴ initially at rest, u = 0)

a = \(\frac{2s}{t^{2}} \) …………… (i)

Again from second law of motion

F = ma …….. (ii)

And electrostatic force,

F = qE ………. (iii)

From equation (i), (ii) and (iii)

ma = qE

m\(\frac{2s}{t^{2}} \) = qE

E = \(\frac{2ms}{qt^{2}} \) = \(\frac{2 × 6.68 × 10^{-27} × 5 × 10^{-2} }{3.2 × 10^{-19} × (2 × 10^{-6})^2} \) = 521.875

∴ E = 522 Nc^{-1}

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