(a) Find the capacitance. C = F (b) A battery with E = 7 Vis now connected acros

Question

(a) Find the capacitance.

C = F    

(b) A battery with E = 7 Vis now connected across the capacitor. How much energy is stored inthe capacitor when fully charged?

U = J    

(c) The dielectric slab is removed and replaced by a slab ofconducting material with identical dimensions. The spacingsa, b, and c are unchanged. Calculate thecapacitance now.

C' = F    

(d) If slabs of dielectric with k' = 2.8 are next inserted, filling the emptyspace between the top plate and conducting slab and the empty spacebetween the lower plate and the conducting slab, what is thecapacitance of the resulting configuration?

C'' = F

A parallel plate capacitor is shown in the diagram in crosssection. A slab of dielectric (k = 4.5) is positioned midway between theplates. The area of the plates and dielectric slab is A =2.8 m2. The spacings are a = c = 1.5cm and b = 1.5 cm.

Explanation / Answer

(a)  the capacitance. (connecting in series) 1/C=(a+c+b/4.5)/(0*A)= >C=7.44*10-10 F (b)energy is stored in the capacitor when fully charged E=1/2*V2*C=1.82*10-8J (c)capacitor, (2 small capacitor connecting in series) C'=0*A/(a+c)=8.26*10-10F (d)capacitance C''=0*A*2.8/(c+a)=2.31*10-9F (d)capacitance C''=0*A*2.8/(c+a)=2.31*10-9F

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