The potential difference ($V$) between two parallel plates can be calculated using the formula:

$V = E \cdot d$

where:

• $E = 7.4 \times 10^4$ V/m (electric field strength)
• $d = 3.25$ cm $= 0.0325$ m (separation distance, converted to meters)

Now, calculating:

$V = (7.4 \times 10^4) \times (0.0325)$

$V = 2405 \text{ V} = 2.405 \text{ kV}$

Final Answer:

The potential difference between the plates is 2.41 kV (rounded to three significant figures).

Would you like a step-by-step breakdown of the calculation?

Related Questions:

1. What happens to the potential difference if the plate separation is doubled?
2. How would the electric field strength change if the potential difference remains constant but the distance is halved?
3. What is the force experienced by a charge of $2 \times 10^{-6}$ C placed between the plates?
4. If the plates are in a vacuum, does the medium affect the electric field strength?
5. How does inserting a dielectric between the plates affect the potential difference?

Tip:

Always convert units to SI before substituting values into equations to avoid calculation errors!