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

Answer:

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

              $a = \frac{u}{t }$

substituting values  

               $a = \frac{100}{2}$

                $a = 50 \ m/s^2$

The electric field is mathematical represented as

             $E = \frac{(a+g)}{Q/m}$

substituting values

              $E = \frac{(50+9.8)}{0.100}$

              $E = 598 \ N/C$