A parallel-plate capacitor that has air between the plates is initially charged

Question

A parallel-plate capacitor that has air between the plates is initially charged by being connected to a battery. With the battery still connected, one of the following changes is The distance between the plates is doubled. A dielectric with a dielectric constant 3 times that of air is inserted, completely filling the space between the plates. Another identical, but initially uncharged, capacitor is placed next to the first so they are touching, effectively doubling the area of each plate. Rank these changes, from largest to smallest, based on the final potential difference across the capacitor. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) Rank these changes, from largest to smallest, based on the final charge stored on the capacitor. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) Rank these changes, from largest to smallest, based on the magnitude of the final electric field within the capacitor. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) Rank these changes, from largest to smallest, based on the final energy stored by the capacitor. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)

Explanation / Answer

a) battery is connected.

so PD across capacitor will not change because it depends on battery.

so 1 = 2 = 3

b) suppose initial capacitance is C.

and C = e0 A / d

now d' = 2d

C1 = C/2

C2 = 3C

C3 = 2C

and Q = CV

1 < 3 < 2

c) E = V / d

so for 1, E1 = E/2

E2 = E

E3 = E

1 < 2 = 3

d) energy = C V^2 / 2

so this will be same as b.

1 < 3 < 2

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