Short Note

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}. How long will it take for the particle to travel a distance of 40 cm?

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#### Solution

Given:

Charge of the particle, q = 2.5 × 10^{−4} C

Initial velocity, u = 0

Electric field intensity, E = 1.2 × 10^{4} N/C

Mass of the particle, m = 1 g = 10^{−3} kg

Distance travelled, s = 40 cm = 4 × 10^{−1} m

Acceleration of the particle,

\[a = \frac{F_e}{m} = \frac{3}{{10}^{- 3}} = 3 \times {10}^3 \text{ m/ s}^2\]

Let t be the time taken by the particle to cover the distance s = 40 cm. Then,

\[s = \frac{1}{2}a t^2 \]

\[ \Rightarrow t = \sqrt{\frac{2s}{a}} = 1 . 63 \times {10}^{- 2} s\]

Concept: Electric Field - Electric Field Due to a System of Charges

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