Question

How far apart are two conducting plates that have an electric field strength of (4.50 X 10^3 V/m) between them, if their potential difference is (15.0 kV)?

a) (3.33 m)
b) (2.50 m)
c) (1.00 m)
d) (4.50 m)

Answer 1

Using the relationship between electric field strength, potential difference, and distance between plates (E = V/d), the distance between two conducting plates with an electric field strength of 4.50 × 10³ V/m and a potential difference of 15.0 kV is calculated as 3.33 meters. Therefore, the correct answer is option (a) 3.33 m.

Step-by-step explanation:

To find the distance between two conducting plates with a given electric field strength and potential difference, we can use the formula that relates these quantities:

E = V/d

where E represents the electric field strength (in volts per meter, V/m), V represents the potential difference (in volts, V), and d represents the distance between the plates (in meters, m).

Given that the electric field strength (E) is 4.50 × 10³ V/m and the potential difference (V) is 15.0 kV (which is 15.0 × 10³ volts), we can rearrange the formula to solve for d:

d = V/E

Substituting the given values:

d = (15.0 × 10³ V) / (4.50 × 10³ V/m)

After canceling out the common power of ten, we get:

d = 15.0 / 4.50 = 3.33 m

Therefore, the correct option in the final answer is (a) 3.33 m.