Figure 16 shows a parallel-plate capacitor of plate area A and plate separation

Question

Figure 16 shows a parallel-plate capacitor of plate area A and plate separation d. A potential difference V is applied between the plates. The battery is then disconnected and a dielectric slab of thickness b and dielectric constant k is placed between the plates as shown. What is the capacitance Cp before the slab is inserted? What free charge appears on the plates? What is the electric held E phi in the gaps between the plates and the dielectric slab? Calculate the electric held E in the dielectric slab. What is the potential difference between the plates after the slab has been introduced? What is the capacitance with the slab in place?

Explanation / Answer

We know that

a)

The capacitance of the capacitor before slab is inserted is given by

C =eoA/d =8.85*10-12*115*10-4m2/1.24*10-2=820.76*10-14F =8.20pF

b)

The free charge appears on the plates is given by

Q =CV =(820.76*10-14F )(85.5)=70995.74*10-14C=709.95pC

c)

E =q/keoA =709.95*10-12/2.61*8.85*10-12*115*10-4=2672.72V/m

d)

When the dielectric is introduced then electricfield decreases by (1/k) times then E =Eo/k =1024.030V/m

e)

The potential diffrence when dielectric is introduced is v =vo/k =85.5/2.61 =32.758V

f)

The capacitance with the slab is given by

C =eoAk/[kb-t(k-1)]

Now substitute all the values to get the required answer

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