The problem in Persian translates roughly to:

Two large parallel flat plates have potentials of +10 V and -2 V, respectively. The distance between the two plates is 20 cm. Determine the magnitude and direction of the electric field.

To solve this, let's use the formula for the electric field $E$ between two parallel plates with a potential difference:

$E = \frac{\Delta V}{d}$

Where:

• $\Delta V$ is the potential difference between the plates.
• $d$ is the distance between the plates.

Solution:

1. Given Values:

   • Potential of Plate 1, $V_1 = +10 \, \text{V}$
   • Potential of Plate 2, $V_2 = -2 \, \text{V}$
   • Distance between plates, $d = 20 \, \text{cm} = 0.2 \, \text{m}$

2. Calculate the Potential Difference $\Delta V$: $\Delta V = V_1 - V_2 = 10 - (-2) = 10 + 2 = 12 \, \text{V}$

3. Calculate the Electric Field $E$: $E = \frac{\Delta V}{d} = \frac{12}{0.2} = 60 \, \text{V/m}$

4. Direction of the Electric Field: The electric field points from the higher potential to the lower potential. Since Plate 1 has a higher potential (+10 V) than Plate 2 (-2 V), the electric field direction is from Plate 1 toward Plate 2.

Final Answer:

• Magnitude of the Electric Field: $E = 60 \, \text{V/m}$
• Direction of the Electric Field: From Plate 1 (10 V) to Plate 2 (-2 V).

Would you like more details or have any questions?

Here are five related questions for further understanding:

1. How would the electric field change if the distance between the plates was reduced?
2. What would happen to the electric field if both plates had the same potential?
3. How is the electric field between parallel plates different from that of point charges?
4. If the plates were not parallel, how would this affect the electric field calculation?
5. What would the electric field be if the potential difference was doubled?

Tip: When dealing with electric fields between parallel plates, remember that the field is typically uniform, meaning it has the same strength and direction at all points between the plates.