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?

Respuesta :

Answer:

Explanation:

From the question we are told that

     The initial velocity is  [tex]u = 100 m/s[/tex]

       The time taken is  [tex]t = 2.0 s[/tex]

       The charge to mass ratio is  [tex]Q/m = 0.100 C/kg[/tex]

       

Generally the acceleration is mathematically evaluated as

              [tex]a = \frac{u}{t }[/tex]

substituting values  

               [tex]a = \frac{100}{2}[/tex]

                [tex]a = 50 \ m/s^2[/tex]

The electric field is mathematical represented as

             [tex]E = \frac{(a+g)}{Q/m}[/tex]

substituting values

              [tex]E = \frac{(50+9.8)}{0.100}[/tex]

              [tex]E = 598 \ N/C[/tex]