The plate areas and plate separations of five parallel plate capacitors are Rank these according to their capacitances, least to greatest.
A) 1, 2, 3, 4, 5
B) 5, 4, 3, 2, 1
C) 5, 3, and 4 tie, then 1, 2
D) 4, 1, and 2 tie, then 5, 3
E) 3, 5, 1 and 2 tie, 1, 4

Question

The plate areas and plate separations of five parallel plate capacitors are  Rank these according to their capacitances, least to greatest.
A) 1, 2, 3, 4, 5 
B) 5, 4, 3, 2, 1 
C) 5, 3, and 4 tie, then 1, 2 
D) 4, 1, and 2 tie, then 5, 3 
E) 3, 5, 1 and 2 tie, 1, 4

Quizplus · Accepted Answer

Capacitance is directly proportional to the plate area and inversely proportional to the plate separation. Therefore, larger plate areas and smaller plate separations result in greater capacitances. We can compare the five capacitors using the given information:
Capacitor 1: Plate area = 1, Plate separation = 2
Capacitor 2: Plate area = 2, Plate separation = 4
Capacitor 3: Plate area = 3, Plate separation = 6
Capacitor 4: Plate area = 4, Plate separation = 1
Capacitor 5: Plate area = 5, Plate separation = 3

Using the formula for capacitance C = εA/d, where ε is the permittivity of free space, we can calculate the capacitances for each capacitor:

C1 = ε(1)/2 = 0.5ε
C2 = ε(2)/4 = 0.5ε
C3 = ε(3)/6 = 0.5ε
C4 = ε(4)/1 = 4ε
C5 = ε(5)/3 = 1.67ε

Therefore, we can see that Capacitor 4 has the greatest capacitance, followed by Capacitor 5, and Capacitors 1, 2, and 3 have the same capacitance. Thus, the correct order from least to greatest capacitance is 4, 1 and 2 (tie), followed by 5, then 3. Option D matches this order.

The Plate Areas and Plate Separations of Five Parallel Plate