A capacitor consisting of two parallel metal plates has a capacitance of 20 microfarads. what parameter would you change and by how much to increase it to 36 microfarads?
The correct answer and explanation is:
To increase the capacitance of a parallel-plate capacitor from 20 microfarads to 36 microfarads, you can modify either the area of the plates, the separation between the plates, or the dielectric material between them. The capacitance of a parallel-plate capacitor is given by the formula: C=ϵ0⋅AdC = \epsilon_0 \cdot \frac{A}{d}
Where:
- CC is the capacitance.
- ϵ0\epsilon_0 is the permittivity of free space (a constant).
- AA is the area of one of the plates.
- dd is the distance between the plates.
For a fixed dielectric material, the primary parameters affecting capacitance are the plate area AA and the separation distance dd.
Increasing the Plate Area (A):
The capacitance is directly proportional to the area of the plates. If you increase the plate area, the capacitance will increase. To increase the capacitance from 20 microfarads to 36 microfarads, you can increase the area by a factor of: Factor=3620=1.8\text{Factor} = \frac{36}{20} = 1.8
This means you would need to increase the plate area by 80% to achieve the desired capacitance, assuming the separation between the plates and the dielectric material remain unchanged.
Decreasing the Plate Separation (d):
Capacitance is inversely proportional to the distance between the plates. If the distance dd is reduced, the capacitance increases. To increase capacitance by 80%, you would need to reduce the distance between the plates by a factor of 11.8≈0.556\frac{1}{1.8} \approx 0.556, meaning the separation should be reduced to about 55.6% of its original value.
Changing the Dielectric Material:
If the dielectric material between the plates is altered, its dielectric constant κ\kappa can affect the capacitance. A material with a higher dielectric constant increases the capacitance. To increase the capacitance from 20 microfarads to 36 microfarads, the dielectric constant must be increased by a factor of: Factor=3620=1.8\text{Factor} = \frac{36}{20} = 1.8
In summary, to achieve the desired increase in capacitance, you can either:
- Increase the area of the plates by 80%.
- Reduce the separation between the plates to about 55.6% of the original distance.
- Use a dielectric material with a dielectric constant 1.8 times greater than the current material.