A parallel-plate capacitor with plate area A and plate separation d has a capacitance of 3
With the gap between the plates unfilled.The gap is then filled with two dielectric materials,one with dielectric constant 3 and the other one with dielectric constant 6.Each slab of dielectric has area A and thickness d/2,the layering of the dielectrics resulting in the gap being completely filled.Which of the following combinations of capacitors will have the same capacitance as the newly filled parallel-plate one?
A) a 9- capacitor and a 18- capacitor in parallel
B) a 9- capacitor and a 18- capacitor in series
C) a 18- capacitor and a 36- capacitor in parallel
D) a 18- capacitor and a 36- capacitor in series
E) a 9- capacitor and a 36- capacitor in parallel

Question

A parallel-plate capacitor with plate area A and plate separation d has a capacitance of 3 
With the gap between the plates unfilled.The gap is then filled with two dielectric materials,one with dielectric constant 3 and the other one with dielectric constant 6.Each slab of dielectric has area A and thickness d/2,the layering of the dielectrics resulting in the gap being completely filled.Which of the following combinations of capacitors will have the same capacitance as the newly filled parallel-plate one?
A) a 9- capacitor and a 18- capacitor in parallel 
B) a 9- capacitor and a 18- capacitor in series 
C) a 18- capacitor and a 36- capacitor in parallel 
D) a 18- capacitor and a 36- capacitor in series 
E) ​a 9- capacitor and a 36- capacitor in parallel

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor can be expressed as C = εA/d, where ε is the permittivity of the material between the plates, A is the area of each plate, and d is the distance between the plates.

Initially, without any dielectric, the capacitance is given as C1 = ε0A/d, where ε0 is the permittivity of free space.

When the dielectric with constant 3 is added, the capacitance becomes C2 = ε1A/d, where ε1 = 3ε0. Similarly, when the dielectric with constant 6 is added on top, the total capacitance becomes C3 = ε2A/d, where ε2 = 6ε0.

The equivalent capacitance of capacitors in series is given by 1/Ceq = 1/C1 + 1/C2, and the equivalent capacitance of capacitors in parallel is given by Ceq = C1 + C2.

Now we can substitute the values of capacitances to find the equivalent capacitance in each choice:

A) Capacitances in parallel: 9ε0A/d + 18ε0A/d = 27ε0A/d. This is not equal to C3, so it is not the answer.

B) Capacitances in series: (1/9ε0A/d + 1/18ε1A/d)^-1 = 10ε0A/d. This is not equal to C3 either.

C) Capacitances in parallel: 18ε0A/d + 36ε0A/d = 54ε0A/d. This is also not equal to C3.

D) Capacitances in series: (1/18ε1A/d + 1/36ε2A/d)^-1 = 9ε0A/d. This is the same as C2, so it is the correct answer.

E) Capacitances in parallel: 9ε0A/d + 36ε0A/d = 45ε0A/d. This is also not equal to C3.

Therefore, the best choice is D where the combination of 18- and 36- capacitors in series will result in the same capacitance as the newly filled parallel-plate capacitor with the dielectrics.

A Parallel-Plate Capacitor with Plate Area a and Plate Separation