Figure P4.56(a) depicts a capacitor consisting of two parallel, conducting plate

Question

Figure P4.56(a) depicts a capacitor consisting of two parallel, conducting plates separated by a distance d. The space between the plates contains two adjacent dielectrics, one with permittivity epsilon 1 and surface area A1 and another with epsilon 2 and A2. The objective of this problem is to show that the capacitance C of the configuration shown in Fig. P4.56(a) is equivalent to two capacitances in parallel, as illustrated in Fig. P4.56(b), with C = C1 + C2, (4.33) where C1 = epsilon 1A1/d, (4.134) C2 = epsilon 2A2/d, (4.135) To this end, proceed as follows: Find the electric fields E1 and E2 in the two dielectric layers. Calculate the energy stored in each section and use the result to calculate C1 and C2. Use the total energy stored in the capacitor to obtain an expression for C. Show that (4.133) is indeed a valid result. Figure P4.56: (a) Capacitor with parallel dielectric section, and (b) equivalent circuit.

Explanation / Answer

a) E = V/d s

o E1=E2 = V/d

b) energy per volume in electric field = 1/2 e E^2

so energy in 1 = (1/2 e1 V^2/d^2 ) A d =1/2 e1 A/d V^2

but we know energy in capacitor = 1/2 C V^2

so C1 = e1 A/d

same thing for C2

c) E total then = 1/2 e1 A/d V^2 + 1/2 e2 A/d V^2

so C = (e1 A/d + e2 A/d) = C1 + C2

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