A B-particle is moving perpendicular to the uniform electric field created by a potential difference of $6 × 10^3 V$ across parallel metal plates separated by a vertical distance of 6 cm. What is the acceleration of the beta particle between the plates? (Mass of electron = $9.1 × 10^{-31}kg$)

$1.76 × 10^{16} ms^{-2}$

The correct answer is Option (2) → $1.76 × 10^{16} ms^{-2}$

The electric field (E) between two parallel plates with a potential difference (V) is -

$E=\frac{V}{d}=\frac{6×10^3}{0.06}=10^5N$

$F=qE$

$=(1.6×10^{-19})×10^5$

$=1.6×10^{-14}N$

Also,

force = ma [Newton 2nd law]

$⇒a=\frac{F}{m}=\frac{1.6×10^{-14}}{9.1×10^{-31}}=1.76×10^{16}m/s^2$