Question

How far apart are two conducting plates that have an electric field strength of 4.25×10

3
V/m between them, if their potential difference is 17.5kV ?

Answer 1

The distance between two conducting plates with an electric field strength of 4.25×10³ V/m and a potential difference of 17.5kV is approximately 4.12 meters, calculated using the relationship d = V/E.

Step-by-step explanation:

The distance between two conducting plates can be calculated using the relationship between electric field strength (E), potential difference (V), and separation distance (d), given by the formula E = V/d. To find the separation distance, we rearrange the formula to d = V/E. Given an electric field strength of 4.25×10³ V/m and a potential difference of 17.5 kV (or 17,500 V), the distance d can be calculated by dividing the potential difference by the electric field strength:

d = V/E = 17,500 V / (4.25×10³ V/m) = 4.1176 m

Hence, the plates are approximately 4.12 meters apart.