Compare with/without dielectric

Electricity and Magnetism Level 3

A parallel plate air-capacitor has capacitance C C . A dielectric slab of dielectric constant K K , whose thickness is half of their air gap between the plates, is now inserted between the plates.
The capacitance of capacitor will now be:

None of these 2 K C K − 1 \dfrac{2KC}{K-1} 2 K C K + 1 \dfrac{2KC}{K+1} K C 2 \dfrac{KC}{2} C ( K + 1 ) K − 1 \dfrac{C(K+1)}{K-1}

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Consider a Parallel plate capacitor having each plate area A & separated by a distance d in air.

C 0 = ϵ 0 A d C_0=\frac{\epsilon_0A}{d}

Now when a dielectric slab of thickness half the given separation and di-electric constant K is introduced, \text{Now when a dielectric slab of thickness half the given separation and di-electric constant K is introduced,}

Let the thickness be t(t<d), the field over distance t is E & E’ over distance (d-t) \text{Let the thickness be t(t<d), the field over distance t is E \& E' over distance (d-t)}

If V is the potential difference between the plates of the capacitor, \text{If V is the potential difference between the plates of the capacitor,}

V = E t + E ′ ( d − t ) V = Et+E'(d-t)

We have E ′ E = K \text{We have }\frac{E'}{E}=K

V = E ′ ( d − t + 1 K ) V = E'(d-t+\frac{1}{K})

Again since E ′ = σ ϵ 0 = q ϵ 0 A \text{Again since }E'=\frac{\sigma}{\epsilon_0}=\frac{q}{\epsilon_0A}

V = q ϵ 0 A ( d − t + 1 K ) V = \frac{q}{\epsilon_0A}(d-t+\frac{1}{K})

Hence the Capacitance is given by : C = q V = ϵ 0 A d − t ( 1 − 1 K ) \text{Hence the Capacitance is given by :} C = \frac{q}{V} = \frac{\epsilon_0A}{d-t(1-\frac{1}{K})}

Now putting t = d 2 , we derive C = 2 K C 0 K + 1 \text{Now putting t = }\frac{d}{2} \text{, we derive } C = \frac{2KC_0}{K+1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I guess, you have worked quite hard on an easy problem. First find out the Capacitance in the Dielectric slab that is 2 K ϵ 0 A d 2K\dfrac{\epsilon_{0} A}{d} .

okay?

Now the rest part is again ϵ 0 A d 2 \dfrac{\epsilon_{0}A}{\dfrac{d}{2}} .

Now we can go for a Series connection of 2 capacitors!

Md Zuhair - 3 years, 1 month ago

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