Question

Two capacitors are identical except that they have different insulators in the gap between the conductors. Capacitor A has an air filled gap while capacitor B uses a material with dielectric constant 3 to keep the conductors apart. The two capacitors are connected in series. What is true of the voltage V across the capacitors, the charge Q on the capacitors, and the electric field E in the gap of the capacitors?

Answer 1

In a series connection, capacitor A with an air-filled gap has a larger voltage compared to capacitor B with a dielectric material, the charge on each capacitor is the same, and the electric field is larger for capacitor A.

Step-by-step explanation:

In a series connection, the charge Q on each capacitor is the same, while the voltage V across the capacitors adds up. The electric field E in the gap between the conductors is inversely proportional to the distance d and the dielectric constant K.

For capacitor A with an air-filled gap, the dielectric constant K is close to 1, so the electric field E is larger compared to capacitor B with a dielectric constant of 3, resulting in a larger voltage V across capacitor A.

Therefore, the voltage V across the capacitors is larger for capacitor A, the charge Q on each capacitor is the same, and the electric field E in the gap is larger for capacitor A compared to capacitor B.

Answer 2

Answer: V1=Q/C and V2=Q/(3*C); Q1=Q2=Q; E2=E1/3

Explanation: In order to explain this problem mwe have to consider the expression for a capacitor given by:

C=Q/V also we have to take into account the capacitor with dielectric constant equal to 3.

As the capacitors are identical abd considering that they are connected in series, the total voltage is equal:

V=V1+V2=(Q1/C1)+(Q2/V2) ; Q1=Q2=Q due to are connected in serie.

V=(Q/C1)+(Q/V2)=Q*((1/C1)+1/C2)

besides we also know that C2=3*C1 then we have:

V2=Q/3*C1 so V2=V1/3

Finally, the electric field for each capacitor is given by:

E1=V1/d where d is the separation between the plates

E2=V2/d as we know that V2=V1/3

E2=V1/3*d

then E2=E1/3