Capacitance in a Real-World Scenario?
A parallel plate capacitor is used in a fire alarm system. The plates are 5 cm apart and have an area of 0.05 m^2. The dielectric constant of the material between the plates is 3. The capacitance of the capacitor is 2 μF. How would you modify the capacitor to increase its capacitance by 50%?
1 Answer
📌 CONCEPT: A parallel plate capacitor is used in a fire alarm system to detect electrical discharges, and its capacitance can be modified to increase its sensitivity.
📐 RULE / FORMULA: The capacitance of a parallel plate capacitor is given by C = (ε0 * κ * A) / d, where C is the capacitance, ε0 is the permittivity of free space, κ is the dielectric constant, A is the area of the plates, and d is the distance between the plates.
💡 WORKED EXAMPLE: If the capacitance needs to be increased by 50%, we can either increase the area of the plates by 50% (A' = 1.5 * A) or decrease the distance between the plates by 50% (d' = 0.5 * d). For example, if the original area is 0.05 m^2, the new area would be 0.075 m^2. Alternatively, if the original distance is 5 cm, the new distance would be 2.5 cm. Using the formula, we can calculate the new capacitance C' = (ε0 * κ * A') / d' = 3 * C.
⚠️ COMMON MISTAKE: Students often forget to consider the effect of changing the dielectric constant or the permittivity of free space on the capacitance. In this case, the dielectric constant remains the same, so we only need to consider the change in area or distance.
21 Sept 26
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