Respuesta :
Answer:
a) Q = [tex]9*10^{-10}[/tex] C; E = [tex]2.03*10^5[/tex] V/m; V = 406.8 V
b) dE/dt = [tex]4.07*10^{11}[/tex] V/(m.s); No
c) [tex]J_d[/tex] = 3.6 [tex]A/m^2[/tex]; Equal
Explanation:
Given parameters are:
Area, A = 5 cm^2
Separation, d = 2 mm
Changing current, [tex]i_c[/tex] = 1.8 mA
At time t = 0 the charge [tex]Q_0[/tex] = 0
a) Here, we are asked to find charge, Q, electric field, E, and potential difference, V at time t = 0.5 [tex]\mu s[/tex]
[tex]Q = i_ct = 1.8*10^{-3}*5*10^{-7} = 9*10^{-10}[/tex] C
[tex]E = \sigma/\epsilon_0 = (Q/A)/\epsilon_0 = (9*10^{-10}/5*10^{-4})/(8.85*10^{-12}) = 2.03*10^5[/tex] [tex]\frac{V}{m}[/tex]
[tex]V = Ed = 2.03*10^{-5}*2*10^{-3} = 406.8[/tex] V
b) [tex]E = (Q/A)/\epsilon_0[/tex]
⇒ [tex]\frac{dE}{dt} = \frac{dQ}{dt} \frac{1}{\epsilon_0 A} = \frac{i_c}{\epsilon_0 A} = \frac{1.8*10^{-3}}{5*10^{-4}*8.85*10^{-12}} = 4.07*10^{11}[/tex] V/(m.s)
No, it is constant that does not vary in time because [tex]i_c[/tex] is constant.
c) the displacement current density, [tex]J_d = \epsilon_0\frac{dE}{dt} = \epsilon_0\frac{i_c}{\epsilon_0 A} = i_c/A[/tex]
⇒ [tex]J_d = 1.8*10^{-3}/(5*10^{-4}) = 3.6[/tex] [tex]A/m^2[/tex]
[tex]i_d =J_dA = 3.6*5*10^{-4} = 1.8*10^{-3}[/tex] A
So, [tex]i_c[/tex] and [tex]i_d[/tex] are equal.
Answer:
[tex]a)[/tex]The charge on the plates[tex]$Q=9 * 10^{-10} C ; E=2.03 * 10^{5} \mathrm{~V} / \mathrm{m} ; \mathrm{V}=406.8 \mathrm{~V}$[/tex]
[tex]b)[/tex]The time rate of change of the electric field between the plates[tex]$\mathrm{dE} / \mathrm{dt}=4.07 * 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{s}) ; \mathrm{No}$[/tex]
[tex]c)[/tex]The displacement current density jD between the plates[tex]$J_{d}=3.6 \mathrm{~A} / \mathrm{m}^{2}$[/tex]Equals
Explanation:
Given parameters are:
Area, [tex]$A=5cm^{2}[/tex]
Separation, [tex]$\mathrm{d}=2 \mathrm{~mm}$[/tex]
Changing current, [tex]$i_{c}=1.8 \mathrm{~mA}$[/tex]
At time [tex]$\mathrm{t}=0$[/tex] the charge [tex]$Q_{0}=0$[/tex]
a) Here, we are asked to find charge, [tex]$Q$[/tex], electric field, [tex]$E$[/tex] and potential difference, [tex]$\mathrm{V}$[/tex] at time [tex]$\mathrm{t}=0.5 \mu \mathrm{s}$[/tex]
[tex]$Q=i_{c} t=1.8 \times10^{-3} \times 5 \times10^{-7}=9 \times10^{-10} \mathrm{C}$[/tex]
[tex]$E=\sigma / \epsilon_{0}=(Q / A) / \epsilon_{0}=\left(9 \times10^{-10} / 5 \times10^{-4}\right) /\left(8.85\times 10^{-12}\right)=2.03 \times10^{5} \frac{\mathrm{V}}{m}$[/tex]
[tex]$V=E d=2.03 \times10^{-5} \times2\times 10^{-3}=406.8 \mathrm{~V}$[/tex]
b) [tex]$E=(Q / A) / \epsilon_{0}$[/tex]
[tex]$\Rightarrow \frac{d E}{d t}=\frac{d Q}{d t} \frac{1}{\epsilon_{0} A}=\frac{i_{c}}{\epsilon_{0} A}=\frac{1.8 \times10^{-3}}{5\times 10^{-4} \times 8.85 \times 10^{-12}}=4.07 \times 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{S})$[/tex]
No, it is constant that does not vary in time because is constant.
c) the displacement current density, [tex]$J_{d}=\epsilon_{0} \frac{d E}{d t}=\epsilon_{0} \frac{i_{c}}{\epsilon_{0} A}=i_{c} / A$[/tex]
[tex]$\Rightarrow J_{d}=1.8 \times 10^{-3} /\left(5 \times10^{-4}\right)=3.6 \mathrm{~A} / \mathrm{m}^{2}$[/tex]
[tex]$i_{d}=J_{d} A=3.6 \times5 \times10^{-4}=1.8\times 10^{-3} \mathrm{~A}$[/tex]
So, [tex]$i_{c}$[/tex] and [tex]$i_{d}$[/tex] are equal.
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