Question

A positively charged particle initially at rest on the ground accelerates upward to 200m/s in 2.60s . 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?
Express your answer to two significant digits and include the appropriate units. Enter positive value if the electric field is upward and negative value if the electric field is downward

Answer 1

E = 867 N/C, upward.

Step-by-step explanation:

• Assuming no other forces acting on the particle, it must obey Newton's 2nd Law, as follows:

`F_(net) = m*a (1)`

• There are two forces acting on the particle in the vertical direction, one due to the electric field, and the other due to gravity.
• Since the particle is positively charged, assuming that the electric field aims upward, we can write the following expression for the left side of (1):

`F_(net) = q*E - m*g = m*a (2)`

• Since we know that q/m = 0.1 C/kg,
• ⇒ q = 0.1m C/kg
• Replacing this value of q in (2), and simplifying masses, we get:

`F_(net) = 0.1 *E - 9.8 m/s2 = a (3)`

• We can find the value of a, simply applying the definition of acceleration:

`a =(\Delta v)/(\Delta t) = (200m/s)/(2.6s) =76.9 m/s2 (4)`

• Replacing (4) in (3) and solving for E, we get:
• E = 867.2 N/C, upward.