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WBJEEA parallel plate capacitor has a dielectric slab of dielectric constant K between...
A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C1C1 . When the capacitor is charged, the plate area covered by the dielectric gets charge Q1Q1 and the rest of the area gets charge Q2Q2 . The electric field in the dielectric is E1E1 and that in the other portion is E2E2 . Choose the correct option/options, ignoring edge effects.
Solution:
As E=vdE=vd
E1=vdE1=vd
E2=vdE2=vd
(As both parts have some P.D.) upper capacitor have capacitance
C1=Kε0A/d C1=Kε0A/d
Lower capacitor
C2=2ε0A/dC2=2ε0A/d
∴∴ equivalent capacitance
C=(k+2)ϵ0A/dC=(k+2)ϵ0A/d
∴CC1=k+2k∴CC1=k+2k
And E1 :E2=vd:vd=1E1 :E2=vd:vd=1
E1=vdE1=vd
E2=vdE2=vd
(As both parts have some P.D.) upper capacitor have capacitance
C1=Kε0A/d C1=Kε0A/d
Lower capacitor
C2=2ε0A/dC2=2ε0A/d
∴∴ equivalent capacitance
C=(k+2)ϵ0A/dC=(k+2)ϵ0A/d
∴CC1=k+2k∴CC1=k+2k
And E1 :E2=vd:vd=1E1 :E2=vd:vd=1
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