Question

Figure shows a narrow beam of electrons entering with a velocity of 3 × 10⁷ m/s, symmetrically through the space between two parallel horizontal plates `P_1 P_1^'` and `P_2 P_2^'` kept 2 cm apart.

If each plate is 3 cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end `P_2^'`.

Answer 1

Given: Initial velocity (u) = 3 × 10⁷ m/s

Plate separation (d) = 2 cm = 0.02 m

Plate length (l) = 3 cm = 0.03 m

m = 9.1 × 10⁻³¹ kg

e = 1.6 × 10⁻¹⁹ C

Electron enters midway, so vertical displacement:

y = `d/2`

= `0.02/2`

= 0.01 m

Time of travel between plates (t) = `l/u`

= `0.03/(3 xx 10^7)`

= 10⁻⁹ s

Vertical motion (y) = `1/2 at^2`

⇒ 0.01 = `1/2 a(10^-9)^2`

a = `0.02/(10^-18)`

= 2 × 10¹⁶ m/s²

Electric force (F) = ma = eE

E = `(m a)/e`

= `((9.1 xx 10^-31) xx (2 xx 10^16))/(1.6 xx 10^-19)`

E ≈ 1.14 × 10⁵ V/m

Potential difference (V) = Ed

= 1.14 × 10⁵ × 0.02

≈ 2.3 × 10³ V

≈ 2.3 kV