AP Physics 2 Exam Question:

Two parallel plates are placed 0.02 meters apart and have a potential difference of 200 volts between them. Calculate the electric field between the plates and the force experienced by an electron (charge = -1.6 x 10^-19 C, mass = 9.11 x 10^-31 kg) placed between the plates.

Answer:

Step 1: Calculating Electric Field (E) between the Plates

The formula for electric field between parallel plates is given by: [ E = \frac{V}{d} ]

where: E = electric field (in N/C) V = potential difference (in volts) d = distance between the plates (in meters)

Given: V = 200 V, d = 0.02 m

Therefore, the electric field between the plates is 10000 N/C.

Step 2: Calculating Force (F) experienced by an electron between the Plates

The force experienced by a charged particle in an electric field is given by the equation: [ F = qE ]

where: F = force (in N) q = charge of the particle (in C) E = electric field (in N/C)

Given: q = -1.6 x 10^-19 C, E = 10000 N/C

Therefore, the force experienced by the electron placed between the plates is -1.6 x 10^-15 N.

Thus, the electric field between the plates is calculated to be 10000 N/C and the force experienced by the electron is -1.6 x 10^-15 N.