Respuesta :
Answer:
a) Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F , V = 3.4375 V ,
c) U = 54.7 nJ , d) ΔU = 54 nJ,
Explanation:
a) The capacity of a capacitor is defined
C = Q / V
Q = C V
can also be calculated using geometry consideration
C = e or A / d
we reduce to the SI system
A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²
d = 1.53 cm = 1.53 10⁻² m
we substitute
Q = eo A / d V
Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275
Q = 3.9757 10⁻¹⁰ C
let's reduce to pC
Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)
Q = 397.57 pC
when the capacitor is introduced into the water the dielectric constant is different
Q = k Q₀
Q = 80 397.57
Q = 3.18 104 pC
b) Find capacitance and voltage after submerged in water
C = k C₀
C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²
C = 1.157 10⁻¹⁰ F
V = Vo / k
V = 275/80
V = 3.4375 V
c) The stored energy is
U = ½ C V²
U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275²
U = 5.47 10⁻⁸ J
let's reduce to nJ
109 nJ = 1 J
U = 54.7 nJ
d) energy after submerging
U = ½ (kCo) (Vo / k) 2
U = ½ Co Vo2 / k
U = U₀ / k
U = 54.7 / 80 nJ
U = 0.68375 nJ
the energy change is
ΔU = U₀ -U
ΔU = 54.7 - 0.687375
(a) Charge on the plate before immersion, Qi is 5.258 x 10⁻³ pC and the charge after, Qf is 0.421 pC.
(b) The capacitance and potential difference after immersion is 1.157 x 10⁻¹⁰ F and 3.44 V respectively.
(c) The change in energy of the capacitor is 54.02 nJ.
Charge on the plate before immersion
The charge on the plate is calculated as follows;
[tex]Q =\frac{\varepsilon _o A}{Vd} \\\\Q_i = \frac{8.85 \times 10^{-12} \times (25 \times 10^{-4}) }{275\times 0.0153} \\\\Q_i = 5.258 \times 10^{-15} \ C\\\\Q_i = 5.258 \times 10^{-3} pC[/tex]
Charge on the plate after immersion
[tex]Q_f = k Q_i\\\\Q_f = 80 \times 5.258 \times 10^{-3} \ pC= 0.421 \ pC[/tex]
Capacitance and potential difference after immersion
[tex]C = \frac{k\varepsilon _o A}{d} \\\\C = \frac{80 \times 8.85 \times 10^{-12} \times (25\times 10^{-4} )}{0.0153} \\\\C = 1.157 \times 10^{-10} \ F[/tex]
[tex]V = \frac{V_0}{k}\\\\V = \frac{275}{80} \\\\V = 3.44 \ V[/tex]
Change in energy of the capacitor
The initial energy of the capacitor is calculated as follows;
[tex]U_i = \frac{1}{2} CV^2\\\\U_ i = \frac{1}{2} \times (\frac{\varepsilon _o A}{d} )V^2\\\\U_i = \frac{1}{2} \times (\frac{8.85\times 10^{-12} \times 25 \times 10^{-4}}{0.0153} )\times 275^2\\\\U_i = 5.47 \times 10^{-8} \ J\\\\U_i = 54.7 \ nJ[/tex]
The final energy of the capacitor is calculated as follows;
[tex]U_f = \frac{1}{2} (kC) \times (\frac{V}{k} )^2\\\\U_f = \frac{1}{2} C\times \frac{V^2}{k} \\\\U_f = \frac{1}{k} (\frac{1}{2} CV^2)\\\\U_f = \frac{U_i}{k} \\\\U_f = \frac{54.7 \ nJ}{80} \\\\U_f = 0.68 \ nJ[/tex]
Change in energy is calculated as follows;
[tex]\Delta U = U_i - U_f \\\\\Delta U = 54.7 \ nJ \ - \ 0.68 \ nJ\\\\\Delta U = 54.02 \ nJ[/tex]
Learn more about energy stored in a capacitor here: https://brainly.com/question/13578522
