A parallel plate capacitor has a capacitance of ‘C’. If the distance between the plates is reduced by half and the space between the plates is filled with a medium having dielectric constant 6, the new capacitance is:
12C
Understanding the behavior of a parallel plate capacitor is fundamental in physics, especially when changes occur to its physical dimensions or the material between its plates. This question asks us to determine the new capacitance of a parallel plate capacitor when its plate distance is halved and a dielectric medium is introduced.
Capacitance of a Parallel Plate Capacitor
The capacitance of a parallel plate capacitor depends on several factors: the area of its plates, the distance between them, and the type of material (dielectric) filling the space between the plates. The formula for the capacitance (\(C\)) of a parallel plate capacitor in a vacuum or air is given by:
\[C = \frac{\epsilon_0 A}{d}\]
- Here, \(A\) is the area of each plate.
- \(d\) is the distance between the plates.
- \(\epsilon_0\) (epsilon naught) is the permittivity of free space, a constant value.
Initial Capacitance Setup
According to the question, the initial capacitance of the parallel plate capacitor is \(C\). We can represent this using the formula above:
\[C_{\text{initial}} = C = \frac{\epsilon_0 A}{d_{\text{initial}}}\]
Let's assume the initial distance between the plates is \(d\).
Parameters for New Capacitance
The problem states two changes are made to the parallel plate capacitor:
- The distance between the plates is reduced by half.
- The space between the plates is filled with a medium having a dielectric constant of 6.
Let's denote the new parameters:
- New distance between plates, \(d_{\text{new}} = \frac{d_{\text{initial}}}{2} = \frac{d}{2}\).
- Dielectric constant of the new medium, \(k = 6\).
Calculating New Capacitance
When a dielectric medium with dielectric constant \(k\) is introduced between the plates of a capacitor, the capacitance increases by a factor of \(k\). The formula for the new capacitance (\(C_{\text{new}}\)) becomes:
\[C_{\text{new}} = \frac{k \epsilon_0 A}{d_{\text{new}}}\]
Now, substitute the new distance \(d_{\text{new}} = \frac{d}{2}\) and the dielectric constant \(k = 6\) into the formula:
\[C_{\text{new}} = \frac{6 \epsilon_0 A}{\frac{d}{2}}\]
To simplify the expression, we can bring the denominator's denominator to the numerator:
\[C_{\text{new}} = 6 \times 2 \times \frac{\epsilon_0 A}{d}\]
\[C_{\text{new}} = 12 \times \frac{\epsilon_0 A}{d}\]
Relating to Original Capacitance
We know that the original capacitance \(C\) was defined as \(C = \frac{\epsilon_0 A}{d}\). We can substitute this into our expression for \(C_{\text{new}}\):
\[C_{\text{new}} = 12 \times C\]
Thus, the new capacitance is \(12C\).
Summary of Changes and Impact
Let's summarize the effect of each change on the capacitance:
| Change | Effect on Capacitance Formula | Impact on Capacitance |
|---|---|---|
| Distance between plates reduced by half (\(d \to d/2\)) | \(C \propto 1/d\) | Capacitance doubles (\(\times 2\)) |
| Medium with dielectric constant \(k=6\) introduced | \(C \propto k\) | Capacitance increases by a factor of 6 (\(\times 6\)) |
The combined effect is \(2 \times 6 = 12\). Therefore, the new capacitance is \(12\) times the original capacitance.
The final calculated new capacitance is \(12C\).
Important Questions from Capacitance
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