The electrostatic force (F) between two plates of a charged capacitor can be expressed mathematically using the relationship described in your question. Since it is directly proportional to the product of the charges $q_1$ and $q_2$ on each plate and inversely proportional to the square of the distance $d$ between the plates, the equation can be written as:

$F \propto \frac{q_1 q_2}{d^2}$

To turn this proportionality into an equation, we introduce a constant of proportionality, $k$, which depends on factors such as the medium between the plates (e.g., vacuum, air, or another dielectric material). The complete equation becomes:

$F = k \frac{q_1 q_2}{d^2}$

Where:

• $F$ is the electrostatic force between the plates,
• $q_1$ and $q_2$ are the charges on the plates,
• $d$ is the distance between the plates,
• $k$ is the electrostatic constant, also called Coulomb's constant, which in a vacuum has the value $k \approx 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How does the electrostatic force change if the distance between the plates is doubled?
2. What happens to the force if the charges on both plates are increased by a factor of three?
3. How is the electrostatic constant $k$ affected by different materials between the plates (dielectric effect)?
4. How does the equation for electrostatic force relate to Coulomb’s Law for point charges?
5. Can the force between capacitor plates be altered by changing the geometry of the plates?

Tip: The electric field between parallel capacitor plates is uniform in the region between the plates, but it weakens at the edges.