Answer to Question #215602 in Electricity and Magnetism for NICKO

Question #215602

A parallel plate capacitor has the space between the plates filled with two slabs of dielectric, one with constant κ1 and one with constant κ2. Each slab has thickness d/2, where d is the plate separation. Show that the capacitance is C = 2ε0A /d (κ1κ2/κ1 + κ2)


1
Expert's answer
2021-07-12T12:13:18-0400


For series combination the battery gives same charge and finally the same amount (Q) of charge is taken out to the negative terminal of the battery.

For series capacitance charge is the same, but the potential difference differ by QC\frac{Q}{C}

V=QCV=\frac{Q}{C}

As the question has a dielectric slab, so we take

C=kε0AdC = \frac{kε_0A}{d}

For series combination

V=1C=1C1+1C2C1=kε0A1dC2=kε0A2d1C=1C1+1C2=kε0A1d+kε0A2dCcf=2ε0Ad(k1k2k1+k2)V= \frac{1}{C}=\frac{1}{C_1}+ \frac{1}{C_2} \\ C_1= \frac{kε_0A_1}{d} \\ C_2= \frac{kε_0A_2}{d} \\ \frac{1}{C}=\frac{1}{C_1}+ \frac{1}{C_2} = \frac{kε_0A_1}{d} + \frac{kε_0A_2}{d} \\ C_{cf}= \frac{2ε_0A}{d}(\frac{k_1k_2}{k_1+k_2})


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