Capacitance of a parallel plate capacitor becomes 5/6 times its original value, if a dielectric slab of thickness t = d/3 is inserted between the plates (d is the original separation between the plates) what will be the value of dielectric constant of the slab ?
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Value of the dielectric constant of the slab is 0.625.
Formula for insertion of a dielectric insulator inside a capacitor is:
A e / (d - t + t / k)
Where A, e, d and k take their usual meaning, 't' refers to thickness of inserted dielectric slab.
Substituting values and solving the equation to get 5/6 times the initial capacitance, we get a linear equation in terms of k.
18 k = 10 k + 5
8 k = 5
k = 0.625
Formula for insertion of a dielectric insulator inside a capacitor is:
A e / (d - t + t / k)
Where A, e, d and k take their usual meaning, 't' refers to thickness of inserted dielectric slab.
Substituting values and solving the equation to get 5/6 times the initial capacitance, we get a linear equation in terms of k.
18 k = 10 k + 5
8 k = 5
k = 0.625
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