Question

Consider two parallel plate capacitors. The plates on capacitor B have half the area as the plates on capacitor A, and the plates in capacitor B are separated by twice the separation of the plates of capacitor A. If capacitor A has a capacitance of 69.0 nF, what is the capacitance of capacitor B?

Answer 1

Capacitor B, with half the plate area and twice the separation distance compared to capacitor A, has a capacitance of 17.25 nF, which is a quarter of the 69.0 nF capacitance of capacitor A.

Step-by-step explanation:

To determine the capacitance of capacitor B given the known capacitance of capacitor A (69.0 nF), we need to understand how changes in plate area and separation distance affect capacitance. Capacitance C of a parallel-plate capacitor can be calculated using the formula C = ε0 (A/d), where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.

Given that capacitor B has half the area (A/2) and twice the separation (2d) compared to capacitor A, we can derive its capacitance as follows:

Capacitance of capacitor A: CA = ε0 (A/d)

Capacitance of capacitor B: CB = ε0 ((A/2)/(2d)) = ε0 (A/(4d)) = CA / 4

Therefore, the capacitance of capacitor B is a quarter of the capacitance of capacitor A, which is 17.25 nF.