Physics: Capacitor plates have area 5.00 cm^2 and separation 2.00 mm. the plates are in a vacuum. the charging current i_c has a constant value of 1.80 ma. at t = 0 the charge on the plates is zero. (a) calculate the charge on the plates, the electric field between the plates, and the potential difference between the plates when t = 0.500 ms. (b) calculate de/dt, the time rate of change of the electric field between the plates. does de/dt vary in time? (c) calculate the displacement current density jd between

Capacitor plates have area 5.00 cm^2 and separation

Ответ:

jasonweber8718

13.01.2022

Physics

a) Q = $9*10^{-10}$ C; E = 2.03*10^5 V/m; V = 406.8 V

b) dE/dt = $4.07*10^{11}$ V/(m.s); No

c) J_d = 3.6 A/m^2 ; Equal

Explanation:

Given parameters are:

Area, A = 5 cm^2

Separation, d = 2 mm

Changing current, i_c = 1.8 mA

At time t = 0 the charge Q_0 = 0

a) Here, we are asked to find charge, Q, electric field, E, and potential difference, V at time t = 0.5 $\mu s$

$Q = i_ct = 1.8*10^{-3}*5*10^{-7} = 9*10^{-10}$ C

$E = \sigma/\epsilon_0 = (Q/A)/\epsilon_0 = (9*10^{-10}/5*10^{-4})/(8.85*10^{-12}) = 2.03*10^5$ $\frac{V}{m}$

$V = Ed = 2.03*10^{-5}*2*10^{-3} = 406.8$ V

b) $E = (Q/A)/\epsilon_0$

⇒ $\frac{dE}{dt} = \frac{dQ}{dt} \frac{1}{\epsilon_0 A} = \frac{i_c}{\epsilon_0 A} = \frac{1.8*10^{-3}}{5*10^{-4}*8.85*10^{-12}} = 4.07*10^{11}$ V/(m.s)

No, it is constant that does not vary in time because i_c is constant.

c) the displacement current density, $J_d = \epsilon_0\frac{dE}{dt} = \epsilon_0\frac{i_c}{\epsilon_0 A} = i_c/A$

⇒ $J_d = 1.8*10^{-3}/(5*10^{-4}) = 3.6$ A/m^2

$i_d =J_dA = 3.6*5*10^{-4} = 1.8*10^{-3}$ A

So, i_c and i_d are equal.

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Ответ:

alonzoyt0

13.01.2022

Physics

The charge on the plates $Q=9 * 10^{-10} C ; E=2.03 * 10^{5} \mathrm{~V} / \mathrm{m} ; \mathrm{V}=406.8 \mathrm{~V}$

The time rate of change of the electric field between the plates $\mathrm{dE} / \mathrm{dt}=4.07 * 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{s}) ; \mathrm{No}$

The displacement current density jD between the plates $J_{d}=3.6 \mathrm{~A} / \mathrm{m}^{2}$ Equals

Explanation:

Given parameters are:

Area, $$A=5cm^{2}$

Separation, $\mathrm{d}=2 \mathrm{~mm}$

Changing current, $i_{c}=1.8 \mathrm{~mA}$

At time $\mathrm{t}=0$ the charge $Q_{0}=0$

a) Here, we are asked to find charge, $Q$ , electric field, $E$ and potential difference, $\mathrm{V}$ at time $\mathrm{t}=0.5 \mu \mathrm{s}$

$Q=i_{c} t=1.8 \times10^{-3} \times 5 \times10^{-7}=9 \times10^{-10} \mathrm{C}$

$E=\sigma / \epsilon_{0}=(Q / A) / \epsilon_{0}=\left(9 \times10^{-10} / 5 \times10^{-4}\right) /\left(8.85\times 10^{-12}\right)=2.03 \times10^{5} \frac{\mathrm{V}}{m}$

$V=E d=2.03 \times10^{-5} \times2\times 10^{-3}=406.8 \mathrm{~V}$

b) $E=(Q / A) / \epsilon_{0}$

$\Rightarrow \frac{d E}{d t}=\frac{d Q}{d t} \frac{1}{\epsilon_{0} A}=\frac{i_{c}}{\epsilon_{0} A}=\frac{1.8 \times10^{-3}}{5\times 10^{-4} \times 8.85 \times 10^{-12}}=4.07 \times 10^{11} \mathrm{~V} /(\mathrm{m} . \mathrm{S})$