A parallel plate capacitor has a capacitance of 5μF. If the distance between the plates is halved, what will be the new capacitance? (2023)
Practice Questions
1 question
Q1
A parallel plate capacitor has a capacitance of 5μF. If the distance between the plates is halved, what will be the new capacitance? (2023)
5μF
10μF
2.5μF
20μF
Capacitance C is inversely proportional to the distance d between the plates. If d is halved, C doubles, so the new capacitance is 10μF.
Questions & Step-by-step Solutions
1 item
Q
Q: A parallel plate capacitor has a capacitance of 5μF. If the distance between the plates is halved, what will be the new capacitance? (2023)
Solution: Capacitance C is inversely proportional to the distance d between the plates. If d is halved, C doubles, so the new capacitance is 10μF.
Steps: 6
Step 1: Understand that capacitance (C) is a measure of a capacitor's ability to store charge.
Step 2: Know that the formula for capacitance of a parallel plate capacitor is C = ε * (A/d), where ε is the permittivity, A is the area of the plates, and d is the distance between the plates.
Step 3: Recognize that capacitance (C) is inversely proportional to the distance (d) between the plates. This means that if the distance decreases, the capacitance increases.
Step 4: The original capacitance is given as 5μF. If the distance (d) is halved, the new distance is d/2.
Step 5: Since capacitance is inversely proportional to distance, if d is halved, the new capacitance will be double the original capacitance.
Step 6: Calculate the new capacitance: New capacitance = 2 * 5μF = 10μF.
