Respuesta :
(a) 12.0 V
In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, [tex]C=5.00 \mu F[/tex], the charged stored on the capacitor at the end of the process is
[tex]Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C[/tex]
When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship
[tex]V=\frac{Q}{C}[/tex]
and since neither Q nor C change, the voltage V remains the same, 12.0 V.
(b) (i) 24.0 V
In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:
[tex]C=\frac{\epsilon_0 \epsilon_r A}{d}[/tex]
where:
[tex]\epsilon_0[/tex] is the permittivity of free space
[tex]\epsilon_r[/tex] is the relative permittivity of the material inside the capacitor
A is the area of the plates
d is the separation between the plates
As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: [tex]C'=\frac{C}{2}[/tex]. The new voltage across the plate is given by
[tex]V'=\frac{Q}{C'}[/tex]
and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is
[tex]V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V[/tex]
So, the new voltage is
[tex]V'=2 (12.0 V)=24.0 V[/tex]
(c) (ii) 3.0 V
The area of each plate of the capacitor is given by:
[tex]A=\pi r^2[/tex]
where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be
[tex]A'=\pi (2r)^2 = 4 \pi r^2 = 4A[/tex]
While the separation between the plate was unchanged (d); so, the new capacitance will be
[tex]C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C[/tex]
So, the capacitance has increased by a factor 4; therefore, the new voltage is
[tex]V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}[/tex]
which means
[tex]V'=\frac{12.0 V}{4}=3.0 V[/tex]
The voltage across the capacitor is the ratio of the charge on the plates of the capacitor and its capacitance.
- The voltage across the plates of the capacitors is 12 volts.
- The voltage across the capacitor is 24 volts when the separation between the plates is doubled.
- The voltage across the capacitor is 10 volts when the separation between the plates is unchanged and the area of each plate is doubled.
What is the Voltage across the Capacitor?
The voltage across a capacitor is the maximum amount of voltage that a capacitor can store.
[tex]V = \dfrac {Q}{C}[/tex]
Given that a 5.00μF parallel-plate capacitor is connected to a 12.0 V battery.
Part A:
The charge at the capacitor is given as,
[tex]Q = CV[/tex]
[tex]Q = 5\times 10^{-6}\times 12[/tex]
[tex]Q = 60 \mu\;\rm C[/tex]
The voltage across the capacitor is calculated as,
[tex]V = \dfrac {Q}{C}[/tex]
[tex]V = \dfrac {60\times 10^{-6}}{5\times 10^{-6}}[/tex]
[tex]V = 12\;\rm V[/tex]
Hence the voltage across the plates of the capacitors is 12 volts.
Part B:
The capacitance of the capacitor is given as,
[tex]C = \dfrac {\epsilon _0 \epsilon _m A}{d}[/tex]
Where [tex]\epsilon _0[/tex] is the permittivity of the free space, [tex]\epsilon _m[/tex] is the relative permittivity of the material inside the capacitor, A is the area of the plates and d is the separation between the plates.
- If d = 2d
The capacitance is given as,
[tex]C' = \dfrac {\epsilon _0 \epsilon _m A}{2d}[/tex]
[tex]C' = \dfrac {C}{2}[/tex]
[tex]C' = \dfrac {5\mu}{2}[/tex]
[tex]C' = 2.5 \mu F[/tex]
The voltage across the capacitor will be,
[tex]V' = \dfrac {Q}{C'}[/tex]
[tex]V' = \dfrac {60\times 10^{-6}}{2.5\times 10^{-6}}[/tex]
[tex]V' = 24 \;\rm V[/tex]
Hence we can conclude that the voltage across the capacitor is 24 volts when the separation between the plates is doubled.
- The radius of each plate is doubled and d = unchanged.
The area of the plates is given as,
[tex]A" = 2A[/tex]
The capacitance will be,
[tex]C " = \dfrac {\epsilon _0 \epsilon _m 2A}{d}[/tex]
[tex]C" = 2C[/tex]
[tex]C" = 2\times 5 \mu[/tex]
[tex]C" = 10\mu F[/tex]
The voltage is given as,
[tex]V" = \dfrac {Q}{C"}[/tex]
[tex]V" = \dfrac {60\mu}{10\mu}[/tex]
[tex]V" = 6\;\rm V[/tex]
Hence we can conclude that the voltage across the capacitor is 10 volts when the separation between the plates is unchanged and the area of each plate is doubled.
To know more about the voltage and capacitance, follow the link given below.
https://brainly.com/question/4544702.
