Final answer:
The potential at point B, 4.0 cm to the right of point A, in an electric field of 7.50×104 V/m across plates separated by 4.00 cm, is 3000 V (or 3.0×103 V). None of the provided options match this calculated value, indicating a possible error in the question or options.
Step-by-step explanation:
The given question relates to the concept of electric potential in the context of parallel conducting plates with an electric field present between them. According to the problem statement, the electric field strength between the plates is 7.50×104 V/m, and the separation between the plates is 4.00 cm. To find the potential at point B, which is 4.0 cm to the right of point A (assuming that point A corresponds to the plate with the lowest potential, which is at zero volts), we can calculate the potential difference between the plates first.
The potential difference (V) in an electric field (E) over a distance (d) is calculated using the equation V = E × d. Here, E = 7.50×104 V/m, and d = 4.00 cm, which needs to be converted to meters (d = 0.04 m). Thus, the potential difference between the plates is V = 7.50×104 V/m × 0.04 m = 3000 V. Now, since point B is at the same position as the higher potential plate, we can conclude that point B also has this potential. The potential at point B is therefore 3000 V, or 3.0×103 V, which is not one of the options provided (A) 12.3 V, (B) 8.7 V, (C) 15.6 V, or (D) 10.2 V. Therefore, it seems there might be an error in the question or in the options given.