The question is about two large, parallel flat plates with different electric potentials. The plates are placed with potentials of $+10$ volts and $-2$ volts, and the distance between them is $20$ cm (or $0.2$ meters). We need to determine the magnitude and direction of the electric field between these two plates.

Solution:

1. Electric Field Calculation: The electric field $E$ between two parallel plates with different potentials is given by the formula: $E = \frac{\Delta V}{d}$ where:

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

2. Calculating Potential Difference: $\Delta V = V_{\text{plate1}} - V_{\text{plate2}} = 10 - (-2) = 10 + 2 = 12 \text{ volts}$

3. Distance: Given $d = 20$ cm, we convert it to meters: $d = 20 \, \text{cm} = 0.2 \, \text{m}$

4. Electric Field Calculation: Substitute the values into the formula: $E = \frac{12}{0.2} = 60 \, \text{V/m}$

5. Direction of the Electric Field: The electric field direction is from the higher potential plate ($+10$ volts) to the lower potential plate ($-2$ volts). Thus, it points from the $+10$ V plate to the $-2$ V plate.

Final Answer:

• Magnitude of Electric Field: $60 \, \text{V/m}$
• Direction: From the $+10$ V plate to the $-2$ V plate.

Would you like further details or have any questions?

Here are some related questions:

1. How does changing the distance between the plates affect the electric field?
2. What would happen if both plates had the same potential?
3. How would the electric field change if the potential difference was reversed?
4. Can this setup create a uniform electric field in other configurations?
5. How would the presence of a dielectric material between the plates affect the field?

Tip: Remember, the electric field is stronger when the potential difference is higher or the plates are closer.