Respuesta :
Answer:
a) The ratio is [tex]\frac{U}{U_{0}} =\frac{2}{(1+\frac{1}{\kappa})}[/tex].
b) The total stored energy increases after the dielectric is inserted.
Explanation:
Before the dielectric is inserted:
We have two parallel plate capacitors ([tex]C_{1}[/tex] and [tex]C_{2}[/tex]), each with a capacitance C, connected in series to a battery that has voltage V.
The two capacitors can be replaced by an equivalent one. The equivalent capacitance for two capacitors in series is
[tex]\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{2}{C}[/tex]
The potential differences across the two capacitors are:
[tex]\Delta V_{1}=\frac{Q}{C_{1}}=\frac{Q}{C}[/tex] and [tex]\Delta V_{2}=\frac{Q}{C_{2}}=\frac{Q}{C}[/tex]
The total potential difference [tex]\Delta V[/tex] is the sum of the potential differences across each capacitor:
[tex]\Delta V=\Delta V_{1}+ \Delta V_{2}=2\frac{Q}{C}[/tex]
The initial stored energy [tex]U_{0}[/tex] is:
[tex]U_{0}=\frac{1}{2}C_{eq} \Delta V^{2}=\frac{1}{2}\frac{C}{2}(2\frac{Q}{C})^{2} \ \Longrightarrow\ U_{0}=\frac{Q^{2}}{C}[/tex]
Using [tex]\Delta V=V[/tex] instead we have that [tex]U_{0}=\frac{1}{4}CV^{2}[/tex].
After the dielectric is inserted:
A dielectric with [tex]\kappa > 1[/tex] is inserted between the plates of one of the capacitors while the two capacitors are connected to the battery. The total potential difference reamins the same. If Q is the charge in the plates without the dielectric, in the capacitor with the dielectric the charge increases to [tex]\kappa Q[/tex] and the capacitance to [tex]\kappa C[/tex].
Now we have that the equivalent capacitance is:
[tex]\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{\kappa C}=\frac{1}{C} (1+\frac{1}{\kappa})[/tex]
So the stored energy after the dielectric is inserted [tex]U[/tex] becomes:
[tex]U=\frac{1}{2}C_{eq}\Delta V^{2}=\frac{1}{2} \frac{C}{(1+\frac{1}{\kappa})}V^{2} \Longrightarrow\ U=\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}[/tex]
- So, the ratio [tex]\frac{U}{U_{0}}[/tex] is:
[tex]\frac{U}{U_{0}}=\frac{\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}}{\frac{1}{4}CV^{2}} \Longrightarrow\ \frac{U}{U_{0}}=\frac{2}{(1+\frac{1}{\kappa})}[/tex]
- Because [tex]\frac{U}{U_{0}} >1[/tex] if [tex]\kappa >1[/tex] this means that the total stored energy increases after the dielectric is inserted.
(a) The required ratio of U/U0 for the given sets of capacitors is 2 / (1 + 1/k).
(b) The total stored energy increases after the dielectric is inserted.
Given data:
The capacitance of air-filled parallel plate capacitors are, C1 and C2.
The net capacitance is, C.
The battery voltage is, V.
The total energy stored in the two capacitors without dielectric is, U0.
The total energy stored with dielectric is, U.
The given problem is based on the concepts and fundamentals of parallel plate capacitors.
(a)
- Before inserting the dielectric we have equivalent capacitance as,
[tex]\dfrac{2}{C}=\dfrac{1}{C1}+\dfrac{1}{C2}[/tex]
- And the potential across the two capacitors are,
V1 = Q/C1
V1 = Q/C
And,
V2 = Q/C2
V2 = Q/C
- Total potential across the each capacitor is,
V = V1 + V2
V = Q/C + Q/C
V = 2Q/C
- the energy stored in the capacitor is,
U0 = 1/2 C × V²
U0 = 1/2 (C/2) × (2Q/C)²
U0 = 1/2 (C/2) × 4Q²/C²
U0 = Q²/C ......................................................(1)
- Now after inserting the dielectric, the capacitance becomes,
[tex]\dfrac{1}{C'}=\dfrac{1}{C}+\dfrac{1}{ k C}\\\\\dfrac{1}{C'}=\dfrac{1}{C} \times (1+\dfrac{1}{ k })[/tex]
- the stored energy after the dielectric is inserted U becomes:
[tex]U = \dfrac{1}{2} \times \dfrac{C \times V^{2}}{(1+1/k)}[/tex] ...............................................(2)
Taking the ratio of equation (2) and (1) as,
U/U0 = 2 / (1 + 1/k)
Thus, we can conclude that the required ratio of U/U0 for the given sets of capacitors is 2 / (1 + 1/k).
(b)
From the previous part we obtained that,
U/U0 = 2 / (1 + 1/k)
Clearly, U/U0 > 1 and also it is given that k > 1. Which significantly means that the total stored energy increases after the dielectric is inserted.
Thus, we can conclude that the total stored energy increases after the dielectric is inserted.
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