Question

A parallel plate capacitor with a dielectric slab with dielectric constant k= 3 filling the space between the plates is charged to potential V and isolated. Then the dielectric slab is drawn out and another dielectric slab of equal thickness but dielectric constant k'= 2 is introduced between the plates. The ratio of the energy stored in the capacitor later to that initially is:

Answer 1

Answer:

Ratio of final energy stored and initial energy stored is 3 : 2

Explanation:

As we know that initially it has dielectric constant k = 3

so initial total charge on the capacitor is given as

$Q = kCV$

initial energy stored in it is given as

$U_i = \frac{1}{2}(kC)V^2$

now it is isolated from the battery so charge is conserved on it

now we can find final total energy on it

$U_f = \frac{Q^2}{2C'}$

new capacitance of the plates is given as

$C' = k'C$

so new energy is given as

$U_f = \frac{(kCV)^2}{2k'C}$

$U_f = \frac{k^2}{2k'}CV^2$

Now the ratio of final energy and initial energy is given as

$U_f : U_i = \frac{k}{k'}$

$U_f : U_i = 3 : 2$

#Learn

Topic : Capacitor

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