Question

A positively charged particle initially at rest on the ground accelerated upward to 100m/s in 2.00s. If the particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform, what are the magnitude and direction of the electric field?

Answer 1

Step-by-step explanation:

From the question we are told that

The initial velocity is
`u = 100 m/s`

The time taken is
`t = 2.0 s`

The charge to mass ratio is
`Q/m = 0.100 C/kg`

Generally the acceleration is mathematically evaluated as

`a = (u)/(t )`

substituting values

`a = (100)/(2)`

`a = 50 \ m/s^2`

The electric field is mathematical represented as

`E = ((a+g))/(Q/m)`

substituting values

`E = ((50+9.8))/(0.100)`

`E = 598 \ N/C`