Suppose you have two parallel conducting plates that are separated by 2.9 meters. What will the electric field strength between the plates be (in N/C) if they have a potential difference of [tex]4.9 \times 10^3 \text{ V}[/tex]?

a) [tex]1.69 \times 10^3 \text{ N/C}[/tex]
b) [tex]1.87 \times 10^3 \text{ N/C}[/tex]
c) [tex]2.17 \times 10^3 \text{ N/C}[/tex]
d) [tex]2.34 \times 10^3 \text{ N/C}[/tex]

To calculate the electric field strength between two parallel plates with a given potential difference and separation, use the formula E = V/d. In this problem, E equals approximately a) 1.69 × 10³ N/C for a potential difference of 4.9 × 10³ V and separation of 2.9 meters.

Explanation:

The problem concerns the relationship between electric field strength, potential difference, and the separation of parallel conducting plates. To find the electric field strength between the plates, we use the formula:

E = V/d,

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

Given that the potential difference V is 4.9 × 10³ V, and the separation distance d is 2.9 meters, we can calculate the electric field strength as follows:

E = (4.9 × 10³ V) / (2.9 m) ≈ 1.69 × 10³ N/C.

Therefore, the correct answer is (a) 1.69 × 10³ N/C.