Q : Consider a parallel plate capacitor with rectangular plates of 5 mm x 10 mm

Question

Q : Consider a parallel plate capacitor with rectangular plates of 5 mm x 10 mm and a gap 0.75 mm. The gap between the plates is filled with air. The capacitor is connected to a battery 10 V.

a) Find the electric field between the parallel plates.

b) Find the surface charge density. Use Gauss’s law.

c) Find the total charge in the plate and the capacitance of the capacitor.

d) A dielectric with a dielectric constant ? = 1725 is inserted between the plates while the battery remains connected. Find the electric field, surface charge density, capacitance and potential energy stored in the capacitor.

THANKS SO MUCH

Explanation / Answer

electric field E is given by ,
E = V/d
= 10 / 0.75 * 10-3
= 1333.33 Volts/meter

using gauss law E = /0
where is the surface charge density

=> = E0
=> = 1333.33 * 8.85 * 10-12
=> = 11799.97 C/m2


C = A0/d
   = (5 * 10 * 10-6 ) * 8.85 * 10-12 / 0.75 * 10-3
   = 590 * 10-14 F

Q = CV
    = 590 * 10-14 * 10
    = 590 * 10-13 C

The capacitor was connected to a battery during the entire process.
The capacitance value of the capacitor increases as the dielectric is inserted between the plates.
The charge on the capacitor plates decreases as the dielectric is inserted between the plates.
The electric field between the plates does not change as the dielectric is inserted between the plates.
The energy storage of the capacitor decreases as the dielectric is inserted between the plates.

C' = kC = 1725 * 590 * 10-14
> C' = 1017750 * 10-14 F

Q' = C'V = 1017750 * 10-14 * 10
=> Q' = 1017750 * 10-13 C

' = E0 = is same as in earlier case

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