Respuesta :
Answer:
The energy stored in this new capacitor is [tex]4.4514\times10^{-9}\ J[/tex]
Explanation:
Suppose, Two parallel plates, each having area A = 2180 cm² are connected to the terminals of a battery of voltage [tex]V_{b}= 6\ V[/tex] as shown. The plates are separated by a distance d = 0.39 cm.
We need to calculate the charge
Using formula of capacitance
[tex]C=\dfrac{Q}{V}[/tex]
[tex]\dfrac{Q}{V}=\dfrac{\epsilon_{0}A}{d}[/tex]
[tex]Q=V\times\dfrac{\epsilon_{0}A}{d}[/tex]
Put the value into the formula
[tex]Q=6\times\dfrac{8.85\times10^{-12}\times2180\times10^{-4}}{0.39\times10^{-2}}[/tex]
[tex]Q=2.968\times10^{-9}\ C[/tex]
The distance between the plates is doubled.
We need to calculate the new capacitance
Using formula of capacitance
[tex]C'=\dfrac{\epsilon_{0}A}{d}[/tex]
Put the value into the formula
[tex]C'=\dfrac{8.85\times10^{-12}\times2180\times10^{-4}}{0.78\times10^{-2}}[/tex]
[tex]C'=2.473\times10^{-10}\ F[/tex]
We need to calculate the energy stored in this new capacitor
Using formula of energy
[tex]U=\dfrac{1}{2}C'V^2[/tex]
Put the value into the formula
[tex]U=\dfrac{1}{2}\times2.473\times10^{-10}\times(6)^2[/tex]
[tex]U=4.4514\times10^{-9}\ J[/tex]
Hence, The energy stored in this new capacitor is [tex]4.4514\times10^{-9}\ J[/tex]
