Final answer:
The magnitude of the electric field is 1000 N/C, and it is directed downward. The force on a proton placed at the same point would be 1.6 × 10⁻¹⁶ N, also directed downward.
Step-by-step explanation:
The student's question involves calculating the magnitude and direction of an electric field and the force acting on a charged particle within that electric field. This pertains to the concept of electric fields in physics, a key topic covered in high school and college education.
Part (a)
The magnitude of the electric field (E) can be calculated using the equation F = qE, where F is the force experienced by the charge and q is the charge of the object.
In the given scenario, a small object carrying a charge (q) of 2.0 × 10-8 C is acted upon by a force (F) of 20.0 nN (or 20.0 × 10⁻⁹ N). Using the equation, we can find E by rearranging the equation to E = F / q.
E = (20.0 × 10⁻⁹ N) / (2.0 × 10⁻⁸ C) = 1000 N/C
Since the force is downward, the direction of the electric field is also downward (assuming the charge is positive, as electric fields point away from positive charges and toward negative charges).
Part (b)
If a proton is placed at the same point, the force acting on it can again be found with F = qE, where q is the charge of the proton (1.6 × 10⁻¹⁹ C).
F = (1.6 × 10⁻¹⁹ C) × (1000 N/C) = 1.6 × 10⁻¹⁶ N
The direction of the force on the proton would be the same as the direction of the electric field, which is downward.