Question

a parallel-plate capacitor of capacitance C and distance d between plates is fully charged from a battery of voltage V, and then disconnected. the plates of the capacitor are slowly pulled to the half of the initial distance (d/2).

a) The charged stored on the capacitor is:

b)The voltage across the capacitor is now:

c)How does the energy stored in the capacitor change after this process:

d) How does the electric field inside the capacitor change after this process:

Answer 1

a)

charge remains constant

b)

for constant charge, capacitance in inversely proportional voltage

since the separation remains half, capacitance becoems twicee of its previous value

now, on increasing capacitance, voltage becomes half of it previous value

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c)

U = 0.5 QV

for constant charge, since voltage becomes twice, energy also becomes twice of its previous value

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d)

E = Q/ eo A

sibce charge is constant, so electric field also remians constant of ita previous value

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