Question

À positively charged particle with mass 3.49 an charge
12 c is initially moving with velocity
5.8m/s
enters in an electric field of strength 150 N/c
which is directed
to words right of the page.
conculate the horizontal distance travelled by
partical with in the field in 2.0 s.​

Answer 1

Given that,

Charge = 12 C

Mass = 3.49 g

Electric field = 150 N/C

Time = 2.0 s

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{qE}{m}$

Where, q = charge

E = electric field

m =mass

Put the value into the formula

$a=\dfrac{12\times150}{3.49\times10^{-3}}$

$a=515759.3\ m/s^2$

$a=515.7\times10^{3}\ m/s^2$

We need to calculate the horizontal distance travelled by the particle

Using formula of speed

$v=\dfrac{d}{t}$

$d=v\times t$

Where, v = velocity

t = time

d = distance

Put the value into the formula

$d=5.8\times2.0$

$d=11.6\ m$

Hence, The horizontal distance travelled by the particle is 11.6 m.