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InterviewSolution
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A capacitor is composed of 2 plates separated by a sheet of insulating material 3 mm thick and of relative permitivity 4. The distance between the plates is increased to allow the insertion of a second sheet of 5 mm thick and of relative permitivity εr . If the equivalent capacitance is one third of the original capacitance. Find the value of εr. |
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Answer» C1(ε0 εrA/d) = k(4/3), where k = ε0A x 10+3 The composite capacitor [with one dielectric of εr1 = 4 and other dielectric of εr2 as relative permitivity has a capacitance of C/3. Two capacitors are effectively in series. Let the second dielectric contribute a capacitor of C2. K.(4/9) = C1C2/(C1 + C2) = (K.(4/3).C2)/(K.(4/3) + C2 This gives C2 = 2/3 K (2/3)K = (ε0εr2A)/(5 x 10-3) Er2 = 10/3.k 1/e0A x 10-3 = 10/3 e0 A × 103 1/e0 A × 10−3 = 10/3 = 3.33 |
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