Explanation
The resultant capacitance is,
1C=14+18+14=58
C=85=1.6μF
Energy stored in capacitor is,
E=12CV2
=12×1.6×10−6×152
=0.8×225×10−6J
=1800erg
Since the capacitors are connected in series so charge flowing through them will be same.
Energy stored in first capacitor of capacitance 0.3μF is given by
E1=Q22C1
E1=Q22×0.3
Energy stored in second capacitor of capacitance 0.6μF is given by
E2=Q22C2
E2=Q22×0.6
Now ratio of energy stored by the condensers is given by
E1E2=21
For the first block
Thickness =d1
Dielectric constant =k1
Capacitance =C1
For second block
Thickness =d2
Dielectric constant =k2
Capacitance =C2
Capacitors C1 and C2 are connected in parallel. So equivalent capacitance is :1Ceq=1C1+1C2
Ceq=C1C2C1+C2........(1)
Capacitance in parallel plate capacitor is given as
C1=k1ε0Ad1&C2=k2ε0Ad2
So, equation (1) becomes
Ceq=k1k2ε0Ak1d2+k2d1.............(2)
Equivalent capacitance for the combination is
C=kε0Ad1+d2.........(2)
From equation (1) and (2)
k1k2ε0Ak1d2+k2d1=kε0Ad1+d2
k=k1k2(d1+d2)k1d2+k2d1
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