Keeping the voltage of charging source constant what would be the percentage change in the energy stored in the parallel plate capacitor if the separation between its plates were to be increased by 10%?
Answers
Answered by
65
Capacitance of Parallel Plate capacitor is
C=ε₀A/d
If separation between plates is decreased by 10%
then new separation,
d1=d-dx10/100=9/10d
Therefore, d1=0.9d-----(1)
C1=ε₀A/d1
=ε₀A/0.9d
=1.11ε₀A/d=1.11 C-------------(2)
As energy stored in a capacitor is U=1/2CV²
Volatge is constant (given)
Percentage change in energy will be,
(U1-U)/Ux100%
=[(U1/U)-1]X 100%]
=[(C1/C)-1]X100%=11%
∴Percentage change in energy will be 11%
C=ε₀A/d
If separation between plates is decreased by 10%
then new separation,
d1=d-dx10/100=9/10d
Therefore, d1=0.9d-----(1)
C1=ε₀A/d1
=ε₀A/0.9d
=1.11ε₀A/d=1.11 C-------------(2)
As energy stored in a capacitor is U=1/2CV²
Volatge is constant (given)
Percentage change in energy will be,
(U1-U)/Ux100%
=[(U1/U)-1]X 100%]
=[(C1/C)-1]X100%=11%
∴Percentage change in energy will be 11%
Similar questions