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Class 12 Tuition > Learn Exercise 2 > A parallel plate capacitor with air between the plates...
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A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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Capacitance between the parallel plates of the capacitor, C = 8 pF Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1 Capacitance, C, is given by the formula, Where, A = Area of each plate = Permittivity of free space If distance between...
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Capacitance between the parallel plates of the capacitor, C = 8 pF Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1 Capacitance, C, is given by the formula, Where, A = Area of each plate = Permittivity of free space If distance between the plates is reduced to half, then new distance, d’ = Dielectric constant of the substance filled in between the plates, = 6 Hence, capacitance of the capacitor becomes Taking ratios of equations (i) and (ii), we obtain Therefore, the capacitance between the plates is 96 pF. read less
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An experienced physics professional for iit-jee & neet

When a dielectric of dielectric constant K is inserted between the plates of the parallel plate capacitor then its capacitance increases K times. Here K=6 and the final distance between plates becomes half of the initial distance. So put the values and get the answer.
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Mechanical engineer with 2 years of teaching experience

initially d, therefore, C= EA/d=8 -(1) now d1=d/2, and dielectric is put, therefore C1=KEA/d1. -(2) here k=6 which is dielectric constant we know d1=d/2 substituting it in (1) C1=6EA/(d/2)=12EA/d. -(3) (3)÷(1) C1÷8=12EA/d÷EA/d C1=96 uF
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initially d, therefore, C= EA/d=8 -(1) now d1=d/2, and dielectric is put, therefore C1=KEA/d1. -(2) here k=6 which is dielectric constant we know d1=d/2 substituting it in (1) C1=6EA/(d/2)=12EA/d. -(3) (3)÷(1) C1÷8=12EA/d÷EA/d C1=96 uF read less
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we know C= ∈0A/d. now d′=d/2 and K= 6. so capacitance will be C′=12*2*∈0A/d=12*8=96 pF
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ELECTRICAL ENGINEERING GRADUATE WITH MORE THAN 6 YRS OF TEACHING EXPERIENCE FROM CLASS 9 TO B.TECH

We know, C=εA/D now, C1/C2= (ε1A1/D1)/(ε2A2/D2) =( 1*A1/D1)/(6*A1/0.5D1) = 1/12 HENCE, C2= 12*C1 =(12*8)pF = 96 pf
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