Two large parallel plates are separated by a distance of 5 cm. The plates have equal but opposite charges that create an electric field in the region between the plate. An α particle ( q = 3.2 × 10-27 kg) is released from the positively charged plate, and it strikes the negatively charged plate 2 × 10-6 second later. Assuming that the electric field between the plates is uniform and perpendicular to the plates. what is the strength of the electric field?

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Solution :

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

User Loaded Image | CSIT GuideCharge of ∝ particle (q) = 3.2 × 10-19c

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

Time taken (t) = 2 × 10-6sec

Electric field between the plates (E) = ?

We have

s = ut + \(\frac{1}{2} \)at2

s = \(\frac{1}{2} \)at2                   (∴ 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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