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The problem involves a parallel plate capacitor, and you're given the following details:

- Voltage between the plates: \( 22.0 \, \text{V} \)
- Distance between the plates: \( 3.0 \, \text{cm} \) (which is \( 0.03 \, \text{m} \))

You need to calculate:
a) The electric field between the plates.
b) The change in potential energy when a charge is moved parallel to the field lines.

### Part (a): Electric Field Between the Plates

The electric field \( E \) between the plates of a capacitor is given by the formula:
\[
E = \frac{V}{d}
\]
Where:
- \( V \) is the voltage across the plates.
- \( d \) is the distance between the plates.

Substituting the given values:
\[
E = \frac{22.0 \, \text{V}}{0.03 \, \text{m}} = 733.33 \, \text{V/m}
\]

### Part (b): Change in Potential Energy

To find the change in potential energy \( \Delta U \) when a charge \( q \) is moved a distance \( \Delta x \) in the direction of the electric field, we use the equation:
\[
\Delta U = q \cdot E \cdot \Delta x
\]

Given:
- Charge \( q = 10.4 \, \text{C} \)
- Distance \( \Delta x = 2.0 \, \text{cm} = 0.02 \, \text{m} \)
- Electric field \( E = 733.33 \, \text{V/m} \)

Substituting the values:
\[
\Delta U = 10.4 \, \text{C} \times 733.33 \, \text{V/m} \times 0.02 \, \text{m}
\]
\[
\Delta U = 152.53 \, \text{J}
\]

So, the change in potential energy is \( 152.53 \, \text{J} \).

---

Would you like any more details or have further questions on this problem? Here are some related questions:

1. What is the relationship between the electric field and force on a charge in the field?
2. How does the capacitance of the capacitor change if the distance between the plates is doubled?
3. What is the potential energy of a charge in a uniform electric field if it is held stationary?
4. How would the electric field change if the voltage across the plates was doubled?
5. How does the potential energy change if the charge is moved perpendicular to the electric field instead of parallel?

**Tip:** When dealing with problems involving electric fields and potential energy, always keep track of the direction of the field and the movement of the charge.

There is a voltage of 22.0 V between the two plates of a parallel plate capacitor. The distance between the plates is 3 cm.

a) Calculate the electric field strength between the plates.
b) A positively charged body with a charge of 10.4 C is moved by 2 cm parallel to the field lines. Calculate the change in its potential energy.

Learn how to calculate the electric field between the plates of a capacitor with 22.0 V voltage and 3 cm distance, and the potential energy change when moving a 10.4 C charge by 2 cm. Suitable for high school physics students.

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