Imagine sliding an infinitesimally thin metallic slab along the bottom face of t
ID: 2306855 • Letter: I
Question
Imagine sliding an infinitesimally thin metallic slab along the bottom face of the dielectric shown in Figure (a). We can model this system as a series combination of two capacitors as shown in Figure (b). One capacitor has a plate separation fd and is filled with a dielectric; the other has a plate separation (1 - f)d and has air between its plates. Evaluate the two capacitances in Figure (b): C_1 = ke_0A/fd and C_2 = e_0A/(1 - f)d Find the equivalent capacitance C for two capacitors combined in series: 1/C = 1/C_1 + 1/C_2 = fd/ke_0A + (1 - f)d/E_0A Invert and substitute for the capacitance without the dielectric, C_0 = E_0A/d: 1/C = fd/kE_0A + k(1 - f)d/kE_0A = f + k(1 - f)/k d/E_0A C = k/f + k(1 - f) E_0A/d C = (Use the following as necessary : k, f and C_0.)Explanation / Answer
By the value of k, f and C0
given
C = [k/(f + k*(1-f))]*e0*A/d
Analysis
A. if f tends to zero
means
C = e0*A/d, which means no sielectric
B. if f tends to 1
means
C = k*eo*A/(f*d)
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