Kayla drew a diagram to compare convex and concave lenses. Which labels belong in the areas marked X, Y, and Z? X: Causes light
1 answer:
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Answer:
E_total = 3 N / A
Explanation:
The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.
E_total = E + E₁
the expression for the field of a point charge is
E₁ = k q₁ / r²
for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative
E_total = E -k q₁ / r₂
we substitute
0 = E - k q₁ / r²
q₁ =
let's calculate
q₁ =
q₁ = 1.78 10⁻⁹ C
now we can calculate the field for position x = 4 m
E_total = 4 - 9 10⁹ 1.78 10⁻⁹ / 4²2
E_total = 3 N / A
For an AC circuit:
I = V/Z
V = AC source voltage, I = total AC current, Z = total impedance
Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.
For a resistor R, inductor L, and capacitor C, their impedances are given by:
= R
R = resistance
= jωL
ω = voltage source angular frequency, L = inductance
= -j/(ωC)
ω = voltage source angular frequency, C = capacitance
Given values:
R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s
Plug in and calculate the impedances:
= 217Ω
= j(220)(0.875) = j192.5Ω
= -j/(220×6.75×10⁻⁶) = -j673.4Ω
Add up the impedances to get the total impedance Z, then convert Z to polar form:
Z = +
+
Z = 217 + j192.5 - j673.4
Z = (217-j480.9)Ω
Z = (527.6∠-1.147)Ω
Back to I = V/Z
Given values:
V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)
Z = (527.6∠-1.147)Ω (from previous computation)
Plug in and solve for I:
I = (30.0∠0+220t)/(527.6∠-1.147)
I = (0.0569∠1.147+220t)A
To get the voltages of each individual component, we'll just multiply I and each of their impedances:
= I×
= I×
= I×
Given values:
I = (0.0569∠1.147+220t)A
= 217Ω = (217∠0)Ω
= j192.5Ω = (192.5∠π/2)Ω
= -j673.4Ω = (673.4∠-π/2)Ω
Plug in and calculate each component's voltage:
= (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V
= (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V
= (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V
Now we have the total and individual voltages as functions of time:
V = (30.0∠0+220t)V
= (12.35∠1.147+220t)V
= (10.95∠2.718+220t)V
= (38.32∠-0.4238+220t)V
Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:
V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V
= 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V
= 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V
= 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V