Question

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero voltsWhat is the electric field strength between them, if the potential 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? .

A) (450 V/m)
B) (900 V/m)
C) (225 V/m)
D) (180 V/m)

Answer 1

The electric field strength between two parallel conducting plates with one at zero volts and a potential of 450 V 8.00 cm away from that plate is 5625 V/m, calculated using the formula E = V/d. However, this answer does not match any of the provided multiple-choice options, indicating a possible issue with the question.

Step-by-step explanation:

The question is asking for the electric field strength between two parallel conducting plates given that one is at zero volts and there is a potential of 450 V at a point 8.00 cm from the zero-volt plate.

To find the electric field strength (E), we use the relationship E = V/d, where V is the potential difference and d is the distance. Because we know that the potential difference at 8 cm is 450 V, and the distance from this point to the zero-volt plate is 8.00 cm, we can calculate E as:

E = 450 V / 0.08 m

Therefore, E = 5625 V/m. However, none of the provided options (A) 450 V/m, (B) 900 V/m, (C) 225 V/m, (D) 180 V/m matches this result, so there may be an error in the question options or a misunderstanding of the concepts.