Question

A particle of mass 1 g and charge 2.5 × 10⁻⁴ C is released from rest in an electric field of 1.2 × 10 ⁴ N C⁻¹. (a) Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis? (b) How long will it take for the particle to travel a distance of 40 cm? (c) What will be the speed of the particle after travelling this distance? (d) How much is the work done by the electric force on the particle during this period?

Answer 1

Given:
Charge of the particle, q = 2.5 × 10⁻⁴ C
Initial velocity, u = 0
Electric field intensity, E = 1.2 × 10⁴ N/C
Mass of the particle, m = 1 g = 10⁻³ kg
Distance travelled, s = 40 cm = 4 × 10⁻¹ m

(a) Electric force,

$$\begin{gathered} F_e=qE \\ \Rightarrow F_e=2.5\times {10}^{-4}\times 1.2\times {10}^4=3\mathrm{N} \end{gathered}$$
Force of gravity,
$$\begin{gathered} F_g=mg \\ \Rightarrow F_g=9.8\times {10}^{-3}\mathrm{N} \end{gathered}$$

(b) Acceleration of the particle,
$a=\frac{F_e}{m}=\frac{3}{{10}^{-3}}=3\times {10}^3\mathrm{m}/{\mathrm{s}}^2$
Let t be the time taken by the particle to cover the distance s = 40 cm. Then,
$$\begin{gathered} s=\frac{1}{2}a{t}^2 \\ \Rightarrow t=\sqrt{\frac{2s}{a}}=1.63\times {10}^{-2}\mathrm{s} \end{gathered}$$

(c) Using the third equation of motion, we get
$$\begin{gathered} v^2=u^2+2as \\ \Rightarrow v^2=0+2\times 3\times {10}^3\times 4\times {10}^{-1} \\ \Rightarrow v=4.9\times 10=49\mathrm{m}/\mathrm{s} \end{gathered}$$

(d) Work done by the electric force,
$$\begin{gathered} W=F_{e}s=3\times 4\times {10}^{-1} \\ =12\times {10}^{-1}\mathrm{J}=1.20\mathrm{J} \end{gathered}$$