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Dafna11 [192]
3 years ago
5

Two identical parallel plate capacitors A and B connected to a battery of V volts w/ the switch S closed. The switch is now open

ed and the free space between the plates of the capacitors is filled w/ a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
Physics
1 answer:
Basile [38]3 years ago
5 0
The electrostatic energy stored in a capacitor with capacitance C_0, with a voltage difference V applied to it, and without dielectric, is given by
U_0 =  \frac{1}{2} C_0 V^2
Now let's assume we fill the space between the two plates of the capacitor with a dielectric with constant k. The new capacitance of the capacitor is
C_k = k C_0
So, the energy stored now is
U_k =  \frac{1}{2}C_k V^2= \frac{1}{2}kC_0 V^2

Therefore, the ratio between the energies stored in the capacitor before and after the introduction of the dielectric is
\frac{U_k}{U_0}= \frac{ \frac{1}{2}kC_0 V^2 }{ \frac{1}{2} C_0 V^2}=  k
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