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Question

A parallel plate capacitor has a plate area of 100m2 and plate separation of 10m. The space between the plates is filled up to a thickness of 5m with a material of dielectric constant 10. The resultant capacitance of the system is 'x'pF. The value of ε0=8.85×10-12Fm-1. The value of 'x' to the nearest integer is _______.


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Solution

Step 1. Given data

Area of the parallel plate, A=100m2

Plate separation, d=10m

Thickness, t=5m

Dielectric constant, k=10

Step 2. Finding the capacitance of parallel plate 1 ,C1 and the capacitance of parallel plate 2, C2

By using the formula of capacitance, C

C=kAε0d

C1=10×100ε05 [Where, for plate 1,d=5m,k=10]

C1=200ε0

C2=100×ε05 [Where, for plate 2, d=5m,k=1]

C2=20ε0

Step 3. Finding the equivalent capacitance, Ceq

Ceq=C1×C2C1+C2

Ceq=200ε0×20ε0200ε0+20ε0

Ceq=18.18ε0

Ceq=18.18×8.85×10-19 [Where, ε0=8.85×10-19Fm-1]

Ceq=160.9×10-12

Ceq=161pF

Comparing the above value with Ceq=xpF, we get

Therefore, x=161pF


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