(a) Find the capacitance. C = F (b) A battery with E = 6 Vis now connected acros

Question

(a) Find the capacitance.

C = F    

(b) A battery with E = 6 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' = 2.6 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 = 4.25) is positioned midway between theplates. The area of the plates and dielectric slab is A =2.6 m2. The spacings are a = c = 0.6cm and b = 0.6 cm.

Explanation / Answer

The dielectric constant of the dielectric slab is k =4.25 The area of the capacitor plates is A = 2.6m2 (a)The capacitance of the parallel plate capacitor is C = (o* A/(d - t * (1 - 1/k)))--------------(1) Here,o= 8.85 * 10-12C2/Nm2,d = (a + c) = (0.6 + 0.6) cm = 1.2 cm= 1.2 * 10-2 m and t = b = 0.6 cm = 0.6 *10-2 m Susbstituting the above values in equation (1),we get C = (8.85 * 10-12 * 2.6/(1.2 * 10-2 -0.6 * 10-2 * (1 - 1/4.25))) or C = 31.04 * 10-10 F = 3.104 * 10-9 F= 3.104 nF (b)The battery connected across the capacitor is E = 6 V The energy stored in the battery when fully charged is U = (1/2) * C * E2 or U = (1/2) * 3.104 * 10-9 * (6)2 or U = 55.87 * 10-9 J (c)The capacitance of the parallel plate capacitor when thedielectric slab is removed and replaced by a slab of conductingmaterial with identical dimensions,then the capacitance of thecapacitor is same. (d)The capacitance of the resulting configuration is C = (o* A/(k' * d)) or C = (8.85 * 10-12 * 2.6/(2.6 * 1.2 *10-2)) or C = 7.375 * 10-10 F = 0.7375 * 10-9 =0.7375 nF or U = 55.87 * 10-9 J (c)The capacitance of the parallel plate capacitor when thedielectric slab is removed and replaced by a slab of conductingmaterial with identical dimensions,then the capacitance of thecapacitor is same. (d)The capacitance of the resulting configuration is C = (o* A/(k' * d)) or C = (8.85 * 10-12 * 2.6/(2.6 * 1.2 *10-2)) or C = 7.375 * 10-10 F = 0.7375 * 10-9 =0.7375 nF

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