Two parallel plates, each having area A = 3557 cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.59 cm. The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.18 cm). What is the energy stored in this new capacitor?

Capacitance,C is equal to epsilon times area of the plate divided by distance. The permittivity of air is 8.84 x 10-12 F/m. Capacitance is equal to 2.66 x10^14 F. The energy is equal to 0.5* C*V^2. Substituting, the energy is equal to 0.5* 2.66 x10^14 F * 6^2 V^2 or equal to 4.80 x10^15 W.

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ibiscollingwood ibiscollingwood · Answer 2 · 2022-02-09T03:46:55+01:00

The energy stored in new capacitor is [tex]4.8*10^{-7}Joules[/tex]

Parallel plate capacitor :

The Capacitance is given as,

[tex]C=\frac{\epsilon A}{d}[/tex]

Where [tex]\epsilon[/tex] is permittivity , A is area of plate and d is distance between plates.

[tex]C=\frac{8.854*10^{-12}*3557 }{0.59}=5.3*10^{-8}F[/tex]

When the separation of the plates is doubled then Capacitance became half of original value.

[tex]C'=\frac{C}{2} =\frac{5.3*10^{-8} }{2} =2.67*10^{-8}F[/tex]

The energy stored in new capacitor is given as,

[tex]E=\frac{1}{2} C'V^{2}\\ \\E=\frac{1}{2} *2.67*10^{-8}*(6)^{2}\\ \\ E=4.8*10^{-7}Joules[/tex]

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