Answer:
(A) Potential difference between plates will be 6 volt
(b) Initial energy stored will be [tex]45.126\times 10^{-7}J[/tex]
Explanation:
It is given area of parallel late capacitor [tex]A=8.50cm^2=8.5\times 10^{-4}m^2[/tex]
Initial voltage [tex]V_0=6volt[/tex]
Distance between plates d = 3 mm = 0.003 m
Capacitance [tex]C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-7}\times 8.5\times 10^{-4}}{0.003}=2.507\times 10^{-7}F[/tex]
So charge on the capacitor [tex]q_0=CV_0=2.507\times 10^{-7}\times 6=15.042\times 10^{-7}C[/tex]
(a) This charge will be constant after disconnection.
After disconnection voltage will be
[tex]V_1=\frac{q_0}{C}=\frac{15.042\times 10^{-7}}{2.507\times 10^{-7}}=6volt[/tex]
(b) Initial energy stored in the capacitor
[tex]U=\frac{1}{2}CV^2=\frac{1}{2}\times 2.507\times 10^{-7}\times 6^2=45.126\times 10^{-7}J[/tex]
