Question

How to derive capacitor formula?

Answer 1

To derive the formula for the capacitance of a capacitor, we need to understand the basic principles of how a capacitor works.

1. Basic Definition of Capacitance

Capacitance is the ability of a capacitor to store charge. It is defined as the ratio of the charge \( Q \) stored on one plate of the capacitor to the potential difference (voltage) \( V \) between the two plates. Mathematically:

\[
C = \frac{Q}{V}
\]

Where:

1. \( C \) is the capacitance in farads (F),
   

1. \( Q \) is the charge in coulombs (C),
   

1. \( V \) is the potential difference in volts (V).
   

2. Electric Field Between Plates

For a parallel plate capacitor, the two plates are parallel to each other, and the area of each plate is \( A \). When a voltage \( V \) is applied between the plates, it creates an electric field \( E \) between the plates.

The electric field between two plates of a capacitor is related to the voltage and the distance between the plates:

\[
E = \frac{V}{d}
\]

Where:

1. \( E \) is the electric field (in volts per meter, V/m),
   

1. \( V \) is the voltage between the plates,
   

1. \( d \) is the distance between the plates.
   

3. Relation Between Electric Field and Charge

The electric field \( E \) between the plates also depends on the charge \( Q \) on the plates. The electric field due to a uniformly charged plate is:

\[
E = \frac{\sigma}{\epsilon_0}
\]

Where:

1. \( \sigma \) is the surface charge density, which is the charge per unit area on the plate: \( \sigma = \frac{Q}{A} \),
   

1. \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)).
   

4. Combine the Relations

Now, substitute the expression for \( \sigma \) into the equation for the electric field:

\[
E = \frac{Q}{A \epsilon_0}
\]

Since we know that \( E = \frac{V}{d} \), we can set the two expressions for \( E \) equal to each other:

\[
\frac{V}{d} = \frac{Q}{A \epsilon_0}
\]

5. Solve for the Capacitance

Rearranging the equation above to solve for \( Q \):

\[
Q = \frac{A \epsilon_0 V}{d}
\]

Now, recall the definition of capacitance:

\[
C = \frac{Q}{V}
\]

Substitute the expression for \( Q \):

\[
C = \frac{A \epsilon_0 V}{d \cdot V}
\]

Simplifying:

\[
C = \frac{A \epsilon_0}{d}
\]

6. Final Formula for Capacitance

So, the capacitance of a parallel plate capacitor is:

\[
C = \frac{\epsilon_0 A}{d}
\]

Where:

1. \( C \) is the capacitance in farads (F),
   

1. \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{F/m} \)),
   

1. \( A \) is the area of each plate (in square meters, \( \text{m}^2 \)),
   

1. \( d \) is the distance between the plates (in meters, m).
   

7. Conclusion

This formula shows that capacitance depends on the area of the plates \( A \), the distance between the plates \( d \), and the material between the plates (through the permittivity \( \epsilon_0 \), or the permittivity of the material if it's not a vacuum).

If the capacitor is placed in a different material (other than vacuum), the formula becomes:

\[
C = \frac{\epsilon A}{d}
\]

Where \( \epsilon \) is the permittivity of the material (equal to \( \epsilon_0 \) for a vacuum and higher for other materials).