Question

A particle (q = 4.0 mC, m = 50 g) has a velocity of 25 m/s in the positive x direction when it first enters a region where the electric field is uniform (60 N/C in the positive y direction). What is the speed of the particle 5.0 s after it enters this region?

Answer 1

Answer:

The speed of the particle is 34.66 m/s.

Explanation:

Given that,

Mass of particle = 50 g

Charge of particle = 4.0 mC

Velocity = 25 m/s

Electric field = 60 N/C

Time = 5.0 sec

We need to calculate the speed in y direction

Using formula of speed

$v=a\times t$

Where, a = acceleration

$v=\dfrac{F}{m}\times t$

$v=\dfrac{qEt}{m}$

Put the value in to the formula

$v=\dfrac{4.0\times10^{-3}\times60\times5.0}{50\times10^{-3}}$

$v=24\ m/s$

We need to calculate the speed of the particle

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{25^2+24^2}$

$v=34.66\ m/s$

Hence, The speed of the particle is 34.66 m/s.