Question

If the area of a capacitor's plates is doubled while the distance between them is halved, how will the final capacitance \(\left(C_{f}\right)\) compare to the original capacitance \(\left(C_{i}\right) ?\) a. \(C_{f}=C_{i}\) b. \(C_{\mathrm{f}}=\frac{1}{2} C_{\mathrm{i}}\) c. \(C_{f}=2 C_{i}\) d. \(C_{f}=4 C_{i}\)

Short answer

The final capacitance is four times the initial capacitance: \(C_f = 4C_i\). Thus, option d is correct.

Step-by-step solution

Understand Capacitance Formula

The capacitance of a parallel plate capacitor is given by the formula: \[C = \frac{\varepsilon_0 A}{d}\] where \(C\) is the capacitance, \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the distance between the plates.

Initial Capacitance

Let the initial capacitance be \(C_i\), the initial area be \(A_i\), and the initial distance be \(d_i\). Thus, \[C_i = \frac{\varepsilon_0 A_i}{d_i}\]

Final Capacitance with Changed Parameters

The area is doubled and the distance is halved, meaning the new area \(A_f = 2A_i\) and the new distance \(d_f = \frac{d_i}{2}\). Hence, \[C_f = \frac{\varepsilon_0 A_f}{d_f} = \frac{\varepsilon_0 (2A_i)}{\frac{d_i}{2}}\]

Simplify the Final Capacitance Expression

Simplify the expression for the final capacitance: \[C_f = \frac{\varepsilon_0 2A_i}{\frac{d_i}{2}} = \frac{2 \varepsilon_0 A_i}{\frac{d_i}{2}} = 4 \frac{\varepsilon_0 A_i}{d_i} = 4C_i\]

Conclusion

The final capacitance \(C_f\) is four times the initial capacitance \(C_i\). Thus, \[C_f = 4C_i\] Answer is option d.

Key concepts

Capacitance formula

Capacitance measures how much electric charge a capacitor can store for a given electric potential. Its formula is: \[C = \frac{\varepsilon_0 A}{d}\]
Here,
\( C \) is the capacitance,
\( \varepsilon_0 \) is the permittivity of free space,
\( A \) is the area of the plates,
and \( d \) is the distance between the plates.
This equation shows that capacitance depends directly on the area of the plates and inversely on the distance between them.

So, if we increase the plate area or decrease the distance between them, the capacitance increases. Conversely, reducing the area or increasing the distance decreases the capacitance.

Parallel plate capacitor

A parallel plate capacitor consists of two conducting plates separated by an insulating material. The plates store equal and opposite charges when connected to a battery.

The capacitance formula for such a capacitor is \( C = \frac{\varepsilon_0 A}{d} \), highlighting two important design aspects:

• Increase in Plate Area: Doubling the area of the plates \( A_f = 2A_i \) will double the capacitance since area and capacitance are directly proportional.
• Decrease in Separation Distance: Halving the distance between the plates \( d_f = \frac{d_i}{2} \) will also double the capacitance because distance and capacitance are inversely proportional.

Together, these changes can greatly increase the final capacitance.
In the given exercise, doubling the area and halving the distance between the plates results in a final capacitance that is four times the initial capacitance.

Electric permittivity

Electric permittivity, represented by \( \varepsilon_0 \), is a measure of how much resistance is encountered when forming an electric field in a medium. For a vacuum, its value is approximately \( 8.85 \times 10^{-12} \; F/m \) (Farads per meter).

Permittivity is crucial in determining the capacitance of a capacitor. The higher the permittivity, the more charge a capacitor can store. This concept helps us understand why different materials used as insulators (also called dielectrics) between capacitor plates affect the capacitance.
For a parallel plate capacitor, the permittivity of the medium between the plates modifies the capacitance formula: \[ C = \frac{\varepsilon A}{d} \] where \( \varepsilon \) is the permittivity of the insulating material.

In summary, the permittivity of the medium in which the capacitor operates directly affects its capacitance.