Question

A parallel plate capacitor has plates of area A = 5.50 ✕ 10−2 m2 separated by distance d = 1.32 ✕ 10−4 m. (The permittivity of free space is ε0 = 8.85 ✕ 10−12 C2/(N · m2).) (a) Calculate the capacitance (in F) if the space between the plates is filled with air. .

What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant κ = 3.10

as in figure (a), and figure (b)? (Hint: One of the capacitors is a parallel combination and the other is a series combination.) Two figures each show a horizontally-oriented parallel-plate capacitor partially filled with a dielectric material with dielectric constant κ. The other half is filled with air. Each plate has area A and a distance d separates the plates. Figure (a): The dielectric material fills the top half of the space between the plates. Figure (b): The dielectric material fills the right half of the space between the plates.

Answer 1

Capacitance of the capacitor is given by formula : _o*A/d

Hence C = 8.85 x 10^-12 * 5.50 x 10^-2/1.32 x 10^-4

C = 36.875 x 10^-10F = 3.68nF

Now, if the space in between in filled as specified in a), there is division of capacitor into two halves of half it's area, hence by the formula we can see that reducing the area to half will reduce the capacitance to half.

Therefore in case a) we can consider as two capacitors of capacitance C/2 are connected in parallel, out of which one is filled with dielectric constant K.

So now using formula for parallel combination of capacitors,

C_{​​​​​​Eq} = C_{​​​​​​1} + C_{​​​​​​2}

C_{​​​​​Eq} = C/2 + KC/2 = (K+1)C/2,. Where is the capacitance calculated above and K is the dielectric constant, when the dielectric is filled in a capacitor, then it's Capacitance gets multiplied by K times.

Hence the equivalent capacitance of the combination is:

C_{​​​​​Eq} = {(3.10 + 1)* 3.68/2} nF= 7.54nF

Case b) if the distance in reduced by two then the capacitance bets multiplied by two

Hence new capacitances of the system are 2KC & 2C

And these two are connected in series ,hence

C_{​​​​​Eq} = C_{​​​​​​1} * C_{​​​​​​2} / ( C_{​​​​​​1} + C_{​​​​​​2}) = KC^{​​​​​2}/2C = KC/2 = 3.1*3.68/2 = 5.70 nF