Question

A parallel-plate air capacitor of area A= 14.0 cm^{2}
and plate separation d= 2.20 mm is charged by a battery to a
voltage 56.0 V. If a dielectric material with κ = 4.00 is inserted
so that it fills the volume between the plates (with the capacitor
still connected to the battery), how much additional charge will
flow from the battery onto the positive plate?

Answer #1

here,

Area = 14 cm^2 = 14 * 10^-4 m^2

sepration , d = 2.2 mm = 2.2 * 10^-3 m

potential difference , V = 56 V

dielectric constant , k = 4

the initial charge , Q1 = area * e0 /d * V

final charge , Q2 = area * e0 * k /d * V

so, the additional charge will flow from the battery onto the positive plate , Q = Q2 - Q1

Q = area * e0 * V /d * ( k - 1)

Q = 14 * 10^-4 * 8.85 * 10^-12 * 56 /( 2.2 * 10^-3) * ( 4 - 1)

**Q = 9.46 * 10^-10 C**

**the additional charge will flow from the battery onto
the positive plate is 9.46 * 10^-10 C**

A
parallel plate capacitor of area A = 30 cm2 and separation d = 5 mm
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k= 4, but the battery disconnected before the dielectric
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1) Find the capacitance of the capacitor.
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3) What is the energy stored in the capacitor?
please explain. Thank you

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