A parallel plate capacitor has a capacitance of 5μF. If the distance between the

Practice Questions

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)
  1. 5μF
  2. 10μF
  3. 2.5μF
  4. 20μF

Questions & Step-by-Step Solutions

A parallel plate capacitor has a capacitance of 5μF. If the distance between the plates is halved, what will be the new capacitance? (2023)
  • 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.
  • Capacitance and Distance Relationship – Capacitance (C) of a parallel plate capacitor is given by the formula C = ε(A/d), where ε is the permittivity, A is the area of the plates, and d is the distance between the plates. This shows that capacitance is inversely proportional to the distance between the plates.
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