## Circuit Lab B/C

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### Re: Circuit Lab B/C

Reviving... what does Coulomb's law state?
The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

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### Re: Circuit Lab B/C

What cases do kirchoff's law not apply?
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### Re: Circuit Lab B/C

What cases do kirchoff's law not apply?
They always apply?

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### Re: Circuit Lab B/C

What cases do kirchoff's law not apply?
They always apply?
No, there are some cases where they don't apply. With KVL, it doesn't apply with a varying magnetic field, and with KCL, it doesn't apply when the voltage source has a very high frequency. Anyways, your turn!
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### Re: Circuit Lab B/C

What cases do kirchoff's law not apply?
They always apply?
No, there are some cases where they don't apply. With KVL, it doesn't apply with a varying magnetic field, and with KCL, it doesn't apply when the voltage source has a very high frequency. Anyways, your turn!
Whoops...

Calculate the electric field strength 1 micrometer from a proton (e=1.6e-19 C).

If a proton is moving east at 300 m/s and experiences an acceleration of 2.6e10 m/s^2 south, what is the strength and direction of the B-field (the mass of a proton is 1.67e-27 kg)?

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### Re: Circuit Lab B/C

What cases do kirchoff's law not apply?
They always apply?
No, there are some cases where they don't apply. With KVL, it doesn't apply with a varying magnetic field, and with KCL, it doesn't apply when the voltage source has a very high frequency. Anyways, your turn!
So this isn't relevant to the next question, but I think it's informative to clarify the statement about KVL not holding. The issue seems a bit more complicated than KVL being invalid in some cases. KVL ostensibly doesn't hold in the situations mentioned above because there are parasitic inductances that come into play that the normal lumped element model of resistors does not take into account. If the model is made more realistic by adding in these inductances and we are careful with how we define voltages and KVL, then indeed, KVL still holds within a varying magnetic field or with high frequency sources since it is dependent on a conservation of energy argument. It ultimately boils down to the model being insufficient to explain everything that is happening in the circuit rather than a problem with KVL.

This issue actually turned into quite a big discussion online which resulted in a very in-depth look at a series of definitions and experiments. A whole lot more can be found at the following places:
ElectroBOOM's Initial Assertion
Followup video
Relevant paper by MIT Professor John Belcher in response

In any case, sorry about the interruption to the question chain. Please carry on.

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### Re: Circuit Lab B/C

They always apply?
No, there are some cases where they don't apply. With KVL, it doesn't apply with a varying magnetic field, and with KCL, it doesn't apply when the voltage source has a very high frequency. Anyways, your turn!
So this isn't relevant to the next question, but I think it's informative to clarify the statement about KVL not holding. The issue seems a bit more complicated than KVL being invalid in some cases. KVL ostensibly doesn't hold in the situations mentioned above because there are parasitic inductances that come into play that the normal lumped element model of resistors does not take into account. If the model is made more realistic by adding in these inductances and we are careful with how we define voltages and KVL, then indeed, KVL still holds within a varying magnetic field or with high frequency sources since it is dependent on a conservation of energy argument. It ultimately boils down to the model being insufficient to explain everything that is happening in the circuit rather than a problem with KVL.

This issue actually turned into quite a big discussion online which resulted in a very in-depth look at a series of definitions and experiments. A whole lot more can be found at the following places:
ElectroBOOM's Initial Assertion
Followup video
Relevant paper by MIT Professor John Belcher in response

In any case, sorry about the interruption to the question chain. Please carry on.
Thanks for the links! Always glad to see people still helping out with Science Olympiad after high school (assuming you did Science Olympiad in high school?)

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### Re: Circuit Lab B/C

They always apply?
No, there are some cases where they don't apply. With KVL, it doesn't apply with a varying magnetic field, and with KCL, it doesn't apply when the voltage source has a very high frequency. Anyways, your turn!
Whoops...

Calculate the electric field strength 1 micrometer from a proton (e=1.6e-19 C).

If a proton is moving east at 300 m/s and experiences an acceleration of 2.6e10 m/s^2 south, what is the strength and direction of the B-field (the mass of a proton is 1.67e-27 kg)?
1)
$\overset{\rightharpoonup}{E} = K\frac{Q}{d^2}$
Given $K = 9 * 10^9$...

$\overset{\rightharpoonup}{E} = 9 * 10^9\frac{1.6*10^{-19}}{(10^{-6})^2}$
$\overset{\rightharpoonup}{E} = 1440 \frac{N}{C}$

2)
$F = Qv\overset{\rightharpoonup}{B}sin\theta$
There are multiple answers here since the angle between the magnetic field and the velocity vector was not given. So I'll make this simple and say they are orthogonal to each other ($\theta = \pi/2$).
Now according to the Right Hand Rule, if the proton is moving east and is experiencing a force south, the magnetic field must be out of the page.
$(1.67*10^{-27})(2.6*10^{10}) = (1.6*10^{-19})(300)\overset{\rightharpoonup}{B}(1)$
$\overset{\rightharpoonup}{B} = 0.904583 T$
pointed out of the paper.
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UTF-8 U+6211 U+662F
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### Re: Circuit Lab B/C

No, there are some cases where they don't apply. With KVL, it doesn't apply with a varying magnetic field, and with KCL, it doesn't apply when the voltage source has a very high frequency. Anyways, your turn!
Whoops...

Calculate the electric field strength 1 micrometer from a proton (e=1.6e-19 C).

If a proton is moving east at 300 m/s and experiences an acceleration of 2.6e10 m/s^2 south, what is the strength and direction of the B-field (the mass of a proton is 1.67e-27 kg)?
1)
$\overset{\rightharpoonup}{E} = K\frac{Q}{d^2}$
Given $K = 9 * 10^9$...

$\overset{\rightharpoonup}{E} = 9 * 10^9\frac{1.6*10^{-19}}{(10^{-6})^2}$
$\overset{\rightharpoonup}{E} = 1440 \frac{N}{C}$

2)
$F = Qv\overset{\rightharpoonup}{B}sin\theta$
There are multiple answers here since the angle between the magnetic field and the velocity vector was not given. So I'll make this simple and say they are orthogonal to each other ($\theta = \pi/2$).
Now according to the Right Hand Rule, if the proton is moving east and is experiencing a force south, the magnetic field must be out of the page.
$(1.67*10^{-27})(2.6*10^{10}) = (1.6*10^{-19})(300)\overset{\rightharpoonup}{B}(1)$
$\overset{\rightharpoonup}{B} = 0.904583 T$
pointed out of the paper.
I probably should've given Coulomb's constant in the problem... But I think east and south are always orthogonal, no? Anyway, your turn!

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### Re: Circuit Lab B/C

Whoops...

Calculate the electric field strength 1 micrometer from a proton (e=1.6e-19 C).

If a proton is moving east at 300 m/s and experiences an acceleration of 2.6e10 m/s^2 south, what is the strength and direction of the B-field (the mass of a proton is 1.67e-27 kg)?
1)
$\overset{\rightharpoonup}{E} = K\frac{Q}{d^2}$
Given $K = 9 * 10^9$...

$\overset{\rightharpoonup}{E} = 9 * 10^9\frac{1.6*10^{-19}}{(10^{-6})^2}$
$\overset{\rightharpoonup}{E} = 1440 \frac{N}{C}$

2)
$F = Qv\overset{\rightharpoonup}{B}sin\theta$
There are multiple answers here since the angle between the magnetic field and the velocity vector was not given. So I'll make this simple and say they are orthogonal to each other ($\theta = \pi/2$).
Now according to the Right Hand Rule, if the proton is moving east and is experiencing a force south, the magnetic field must be out of the page.
$(1.67*10^{-27})(2.6*10^{10}) = (1.6*10^{-19})(300)\overset{\rightharpoonup}{B}(1)$
$\overset{\rightharpoonup}{B} = 0.904583 T$
pointed out of the paper.
I probably should've given Coulomb's constant in the problem... But I think east and south are always orthogonal, no? Anyway, your turn!
Yes you are definitely correct that east and south are orthogonal, but in the case of the Right Hand "Slap" Rule...(my Physics teacher was great )

...the angle is between the magnetic field vector and the velocity vector. Since you gave the force vector and the velocity vector, the magnetic field vector can be pointed where $\theta \in (0, \pi)$. The vector could have been pointing almost west or almost east while also being slightly out of the page, but the magnitude of the magnetic field vector in both cases would be extremely large compared to when it is perpendicular to the velocity vector. Plus, it makes the math a LOT simpler if we assume the magnetic field vector just comes straight out of the page. Mathematically, the force vector is merely just the cross product of the magnetic field and the velocity vector. Either way, I digress...onto the next question!

Oh no! Dr. Doom has just finished building his capacitor! When designing the build he decided to make the plates have an area of 10 m^2 separated by 1 mm. As the dielectric, he used strontium titanate (k = 310). Dr. Doom plans on using his battery with an EMF = 20kV connected to a set of 20 parallel 100 Mohm resistors in series to charge the capacitor.
1) How long will it take to "fully" charge the capacitor?
2) How much charge will the capacitor hold when it reaches maximum voltage?
3) How much energy is stored in the charged capacitor? How much energy is lost during the process of charging the capcitor?
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