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

The speed of the particle is 34.66 m/s.

Step-by-step 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* t`

Where, a = acceleration

`v=(F)/(m)* t`

`v=(qEt)/(m)`

Put the value in to the formula

`v=(4.0*10^(-3)*60*5.0)/(50*10^(-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=√(25^2+24^2)`

`v=34.66\ m/s`

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