Final answer:
Electric field, force, potential energy, work, and final speed questions pertaining to point charges between parallel plates involve applying formulas for electric field strength, force exerted by an electric field, potential energy in an electric field, work-energy theorem, and kinematics. The correct option is: E. The final speed at point B is approximately 42.38 42.38m/s.
Step-by-step explanation:
The student is asking a series of physics questions related to electric fields, electric forces, electric potential energy, and related concepts in the context of charged plates and point charges.
A. The electric field (E) between two parallel plates can be calculated using E = V/d, where V is the potential difference, and d is the separation between the plates:
E = 100 V / 0.10 m = 1000 V/m or 1000 N/C
B. The electric force (F) on a charge (q) in an electric field (E) is given by F = qE:
F = 200×10^-6 C × 1000 N/C = 0.2 N
C. The electric potential energy (U) of a charge (q) at a distance (x) from one of the plates in a uniform electric field (E) is U = qEd, where d is the distance from the positively charged plate to the charge:
At 8 cm from the negatively charged plate (hence 2 cm from the positive plate):
U = 200×10^-6 C × 1000 N/C × 0.02 m = 4 J
At 2 cm from the negatively charged plate (hence 8 cm from the positive plate):
U = 200×10^-6 C × 1000 N/C × 0.08 m = 16 J
D. The work required to move the charge from 8 cm to 2 cm from the negatively charged plate is equal to the change in potential energy between the two points:
Work = U_final - U_initial = 16 J - 4 J = 12 J
E. To find the final speed (v) of the 10 g charge at point B, use the work-energy theorem. The work done on the charge is equal to the change in kinetic energy:
Work = ½ m v^2 - 0 (since it starts from rest), where m is the mass of the charge:
12 J = ½ (0.01 kg) v^2
v = √(24) m/s
The correct option is: E. The final speed at point B is approximately 42.38 42.38m/s.