Physics, 05.03.2020 18:50, kevinseven23

The battery is now disconnected from the plates and the separation of the plates is doubled ( = 0.78 cm). What is the energy stored in this new capacitor?

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answered: brandy127

The energy stored in this new capacitor is $4.4514\times10^{-9}\ J$

Explanation:

Suppose, Two parallel plates, each having area A = 2180 cm² are connected to the terminals of a battery of voltage $V_{b}= 6\ V$ as shown. The plates are separated by a distance d = 0.39 cm.

We need to calculate the charge

Using formula of capacitance

$C=\dfrac{Q}{V}$

$\dfrac{Q}{V}=\dfrac{\epsilon_{0}A}{d}$

$Q=V\times\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$Q=6\times\dfrac{8.85\times10^{-12}\times2180\times10^{-4}}{0.39\times10^{-2}}$

$Q=2.968\times10^{-9}\ C$

The distance between the plates is doubled.

We need to calculate the new capacitance

Using formula of capacitance

$C'=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C'=\dfrac{8.85\times10^{-12}\times2180\times10^{-4}}{0.78\times10^{-2}}$

$C'=2.473\times10^{-10}\ F$

We need to calculate the energy stored in this new capacitor

Using formula of energy

$U=\dfrac{1}{2}C'V^2$

Put the value into the formula

$U=\dfrac{1}{2}\times2.473\times10^{-10}\times(6)^2$

$U=4.4514\times10^{-9}\ J$

Hence, The energy stored in this new capacitor is $4.4514\times10^{-9}\ J$

answered: Guest

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answered: Guest

Did you ever figue this

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The battery is now disconnected from the plates and the separation of the plates is doubled ( = 0.78...