Three parallel-plate capacitors each store the same amount of charge. Capacitor 1 has a plate area Aand a plate separation d. Capacitor 2 has a plate area 2A and a plate separation d. Capacitor 3 has a plate area A and a plate separation 2d.

A. Rank the three capacitors, based on their capacitance.
B. Rank the three capacitors, based on the potential difference between the plates.
C. Rank the three capacitors, based on the electric field magnitude between the plates.
D. Rank the three capacitors, based on the energy stored.
E. Rank the three capacitors, based on the energy density.

Explanation:

A.

The capacitance of a capacitor, C is:

C = £° \(A * d)

where,

A = plate area

d = distance between the plate

£° = free space permitivity

In Capacitor 1, plate area A, plate separation, d

C1 = £°\(A * d)

In Capacitor 2, plate area 2A, plate separation d

C2 = £°\(2A * d)

= 2 * C1

In Capacitor 3, plate area A, plate separation 2d

C3 = £°\(A * 2d)

= 1/2 * C1 = 0.5 * C1

So,

C3 < C1 < C2.

B.

The potential difference between two plates on a capacitor is:

V = Q * C

Where,

V = potential difference

Q = charge on the plate

In Capacitor 1, C = C1

Potential difference, V1 = Q * C1

In Capacitor 2: C2 = 2 * C1

Potential difference, V2 = Q * 2C1

= V1 * 2

In Capacitor 3: C3 = 1/2 * C1

Potential difference: V3 = 1/2 * Q * C1

=2 * V1

So,

V2 < V1 < V3.

C.

The electric field magnitude between the plates of a capacitor:

E = V/d

where,

V = the potential difference between the plates

d = distance between the plates

In Capacitor 1, C1; potential difference V1, plate separation d

electric field, E1 = V1/d

In Capacitor 2, C2; potential difference V2 = 1/2 * V1, plate separation d

electric field, E2 = 1/2 * V1/d

= V1/2d

= 1/2 * E1

In Capacitor 3, C3; potential difference 2V1, plate separation 2d

electric field: E3 = 2 * V1/2d

= V1/d = E1

So,

E2 < E1 and E3.

D.

Energy stored in a capacitor is

W = 1/2 * Q * V

In Capacitor 1, C1; potential difference, V1

energy: W1 = 1/2 * Q * V1

In Capacitor 2, C2; potential difference, V2 = 1/2 * V1

energy: W2 = 1/2 * Q/(V1/2)

= 1/2 * W1

In Capacitor 3, C3; potential difference, V3 = 2 * V1

energy: W3 = 1/2 * Q(2 * V1)

=2 * U1

So,

W2 < W1 < W3.

E.

The energy density in a capacitor is:

u = 1/2 * £° * E^2

where,

E is the electric field strength

In Capacitor 1, C1; electric field, E1

Energy density: u1 = 1/2 * £° * E1^2

In Capacitor 2, C2; electric field, E2 = 1/2 * E1

energy density: u2 = 1/2 * £° (E1/2)^2

= 1/4 * E1

In Capacitor 3, C3; electric field E3 = E1

Energy density: u3 = 1/2 * £° * E1^2

So,

u2 < u1 and u3.

the answer is kevlar

connect the batteries in series and hope