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 6 and dielectric constant k is placed between the plates as shown. Assume What is the capacitance C0 before the slab is inserted? What free charge appears on the plates? What is the electric field E0 in the gaps between the plates and the dielectric slab? Calculate the electric field 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? Fig. 16.

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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