The formula to calculate the force between parallel plate capacitors is:
\[ F = \frac{Q^2}{2 \cdot C_{\parallel}} \]
Where:
Force is the push or pull that occurs between two charged objects, resulting in attraction or repulsion, depending on the nature of the charges.
Charge is a fundamental property of matter that causes objects to experience a force when placed in an electrostatic field.
Parallel plate capacitance is the ability of a system of two parallel plates to store electric charge, commonly used in electronic devices and circuits.
Let's assume the following values:
Using the formula:
\[ F = \frac{0.3^2}{2 \cdot 0.018} = 2.5 \text{ N} \]
The force is 2.5 Newtons.
| Charge (C) | Parallel Plate Capacitance (F) | Force (N) |
|---|---|---|
| 0.1 | 0.01 | 0.50 |
| 0.1 | 0.02 | 0.25 |
| 0.1 | 0.03 | 0.17 |
| 0.1 | 0.04 | 0.13 |
| 0.1 | 0.05 | 0.10 |
| 0.2 | 0.01 | 2.00 |
| 0.2 | 0.02 | 1.00 |
| 0.2 | 0.03 | 0.67 |
| 0.2 | 0.04 | 0.50 |
| 0.2 | 0.05 | 0.40 |
| 0.3 | 0.01 | 4.50 |
| 0.3 | 0.02 | 2.25 |
| 0.3 | 0.03 | 1.50 |
| 0.3 | 0.04 | 1.13 |
| 0.3 | 0.05 | 0.90 |
| 0.4 | 0.01 | 8.00 |
| 0.4 | 0.02 | 4.00 |
| 0.4 | 0.03 | 2.67 |
| 0.4 | 0.04 | 2.00 |
| 0.4 | 0.05 | 1.60 |
| 0.5 | 0.01 | 12.50 |
| 0.5 | 0.02 | 6.25 |
| 0.5 | 0.03 | 4.17 |
| 0.5 | 0.04 | 3.13 |
| 0.5 | 0.05 | 2.50 |