Answer :
Answer:
a) [tex] 3.89\times10^{-3} J[/tex]
b) [tex] 0.01 J [/tex]
c) [tex] 6.11\times10^{-3} J[/tex]
Explanation:
A capacitor is an electronic instrument widely used in circuits and electronic tools. It is usually a pair of parallel plates used to store charge. The energy stored in the capacitor (E) is related with the charge (Q) on the plates and the potential difference across the plates [tex]\Delta V [/tex] as:
[tex]E=\frac{Q\Delta V}{2} [/tex] (1)
Capacitance is the charge over the potential difference (with Faraday (F) units):
[tex]C=\frac{Q}{\Delta V} [/tex]
[tex]Q=C\Delta V [/tex] (2)
Using (2) on (1):
[tex] E= \frac{C(\Delta V)^{2}}{2}[/tex] (2)
That is the energy stored on a capacitor in terms of the potential difference and the capacitance
a) [tex]E= \frac{11.5\times10^{-6}(26.0)^{2}}{2} =3.89\times10^{-3} J[/tex]
The dielectric constant (k) of a material is the defined as the capacitance the capacitor increases due the dielectric (C) over the initial capacitance (Co):
[tex] k=\frac{C}{C_{0}}[/tex]
So, the new capacitance is [tex]C=k*C_0 [/tex]
Using this value on (2)
b) [tex]E= \frac{k*C_0(\Delta V)^{2}}{2}[/tex] [tex]=\frac{3.75*11.5\times10^{-6}(26.0)^{2}}{2}=0.01 J [/tex]
c) Subtracting the result obtained on a) of b) the energy change during the insertion of the dielectric is [tex] 6.11\times10^{-3} J[/tex]