Electric field between widening pair of conductors

The question is: There are two plates of conductors as shown in figure below. Among the four possibilities shown in figure, which one has the electric field lines correctly? 

Here for the answer:

Firstly let us reduce the possibility. In a conductor, there can never be any tangent electric field close to surface of conductor.  So the electric field going in/out must be perpendicular to the surface clost to the conductor. Hence (a) and (b) are not possible.

In (c) and (d), we see that in (c) the electric field is denser in the narrower edge than wider edge of the conductor. (d) is uniformly spread electric field. To make this choice, we play smart, and have the following setup: two capacitor of same area but the distance between the plates are narrower and the other wider. They are connected to wire. Doesn't it look similar to the setup in question, just that we can solve for this.

We can see that the capacitance of the narrower sepration is more than the wider sepration. i.e., \( C_{narrower} > C_{wider} \).

Now we use \( Q=CV \) to figure out which contain more charges. Since they both are at the same potential (as they are connected by the wire), we have:

$$ V=Q_{narrow}/C_{narrow}=Q_{wide}/C_{wide} $$

Since \( C_{narrow} > C_{wide} \) , \( \therefore Q_{narrow} > Q_{wide} \) .  And more Q means more electric field. Hence choice (c) is correct.

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Lenz Law - Find the direction of induced current

 In this problem, you are asked to find the direction of induced current in the loop (shown in Red). The situation is the loop is beside a long wire (shown in dark blue) in which current is increasing in (a) and current is decreasing in (b).

I will give you chance to guess so you need to scroll to see the answer.

Lenz law is one of the toughest part in high school/junior college physics. I still get a bit nervous when student ask me 😓. Some always ask me strange question where I feel not so sure. 😊

Below

is

the 

answer

The answer is both CCW (counter clockwise). Why so? Let's do stepwise analysis:

1. Let us try to find the direction of magnetic field from a long wire. We apply the right hand grip rule, where thumb is in the direction of current and fingers are in the direction of magnetic field. Hope you can figure that it is out of page above the long wire and into the page below the wire.
2. Let us now see the magnetic field inside the loop. 
1. For (a) magnetic field is into the page and increasing due to increase of current.
2. For (b) magnetic field is out of the page and decreasing due to decrease of current
3. Induction Response: According to the Lenz law, the induction is such that it resist the change. So to do that:
1.  For (a) the induced magnetic field (shown in Red) will be against the magnetic field from the wire so that it tries its best to decrease it. Once again. The magnetic field from wire increase, the induction is such that it will oppose it hence reverse direction
2. For (b) the induced magnetic field (shown in Red) will be in same direction of the magnetic field from the wire so that it tries its best to keep the magnetic field the same. Once again. The magnetic field from wire decrease, the induction is such that it will help keep the same so it will be in same direction.
4. Now only focus on the induced magnetic field (shown in Red). We will again apply the same right hand grip rule. But with thumb in direction of magnetic field and fingers in direction of current. (interestingly the right hand grip rule can be applied both ways, thumb in direction of current or thumb in direction of magnetic field). Now for the answer"
1. For (a): Applying the right hand grip rule, with thumb out of the page, the fingers are in CCW (counter clock wise), Hence the current is in counter clockwise direction
2. For (b): exactly the same argument. Hence CCW

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Relativity - Can the order of event be changed - Causality question

Let us consider a rest frame where two events A and B happened at the same point. The event A happens earlier than event B i.e., mathematically,  (\( t_A\) < \(t_B  \)).

The question is: Can you have a frame moving with velocity \( v \) where event B happens earlier than event A i.e.  (\( t_A\) > \(t_B  \))?

(An exam question from a European university where A. Einstein studied and H. Minkowski

 was the professor, and called Einstein a "lazy dog" 😄)

I will answer in three different ways: (1) Melting of ice (2) Lorentz transformations (3) Minkowski diagram

Here is my answer 1, by Gedanken experiment - melting of ice:

Let us have a thought experiment (gedanken experiment) where a ice cube in a glass melts into water. Here we define event A when water exist as ice and event B when all the ice melts. So in rest frame  \( t_A\) < \(t_B  \)). 

Do you think you can have a frame where the event B happens before event A? i.e., water turning into ice cube? Obviously not, it would violate second law of thermodynamics, which says disorder (entropy) should always increase.

Here is a standard answer 2, by Lorentz transformations:

Lorentz transformations for time is:

$$ t'_A = \gamma (t'_A-v x_0/c^2) \quad t'_B = \gamma (t'_B-v x_0/c^2) $$

Where the position \( x_0 \) is same for both events A and B

Subtracting them we get:

$$ t'_A - t'_B = \gamma (t'_A-t'_B) $$

Therefore if \( t_A-t_B < 0 \) implies \( t'_A-t'_B < 0 \). i.e., If event A happens before event B in rest frame at the same place, then the same happens in all other frames. The time interval between A and B changes by factor of \( \gamma \), but the order cannot be changed!

Here is answer 3, by Minkwoski diagram:

Here I will only prove that the order of event A and B can be changed!! if the two event happend at different position. It will be for you to prove that if we graduall make the position closer the order of event cannot be changed. I will take the diagram from this great web site: https://tikz.net/relativity_minkowski_diagram/ that has very many Minkowski diagrams. 

The above diagram shows indeed in some circumstances the order of the event can be can changed in the moving frame w.r.t. a rest frame, but only when two event happens are seperated by fairly large distance. It will never happen if the two events are brought closer so that they happen at the same point as seen in rest frame. 

Hope you have enjoyed the question! 😊

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Money can play tricks with mind

If you want to test a man's character, give him power -- Abraham Lincoln 

This is not a punch line, but something litterally happens. Here is an incident:

Some Indians who reside in western country, play an interesting game. They get a large loan from the bank in the foreign country and use that money to buy property in India. This way, they do not have to pay high interest rate. It seems like a fair game, and there’s nothing wrong with it.

Here is one instance, where a colleague of mine got gamed! He borrowed a large amount of money and transferred to close and trusted relative who can do his paper work and buy the land in India. Once money was transferred and he did the required paper work and purchased the land. Soon enough, he realized that it was not the premium land he wanted to buy. It was a cheaper one, and a good fraction of money his relative kept for himself.

I don’t think his trusted relative was bad; it’s just that he must have never seen so much money in his bank account. When he did, his mind went bonkers. So, never trust anyone with large sums of money, including your close relatives!

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Physics - Why charges accumulate more at the sharp edges in a conductor

A bit about Lightening Conductor:

It is good to know this fact "why charges accumulate more at the sharp edges", as the lightening conductor uses this fact to prevent lightening strike directly to the building. A lightening conductor is a long thick wire from the top of the building to inside the earth. The wire at the building end is kept sharp. 

So during the stormy weather when the clouds and earth are oppositely charged and creates ideal situation for lightening. Since the lightening conductor is connected to earth also becomes charged, the sharp edge has more charges and therefore instense electric field is created that can break electic conductivity of air. It literally creates the lightening from earth and tempts the lightening from clouds to strike the wire. The building is therefore saved.

Now for the physics of Charge Accumulation at the Sharp Edges 

Let me show the figure of a charged conductor below. You can see where the curvature is small (sharp edges), the accumulation of the charge is more. Correspondingly, where the curvature is large (flatter portion) the charges per unit area  is less.

Defining the Problem:

So let us try to define the problem more concretely so we can prove it. Here is the Problem to prove:  The charge density is more at high curvature then low curvature area.

Strange Properties of Conductor:

Since are are talking about conductor, let me elaborate some interesting properties of the conductor:

1. For a conductor, the potential at any two points in a conductor is the same. (If it is not, the electrons will move quickly to make it so!)
2. In a conductor, there are no charges inside the conductor. All the charges are on the surface! Strange, but it is because if there are, say positive charges, inside the conductor, the electron nearby will come to neutralize. It will do to keep the potential between two points the same)
3. The electric field comming out of surface of the conductor is alway perpendicular to surface. (for positive charged conductor it is comming out, for negatively charged conductor it is going in. But always perpendicular to surface)
4. More charge accumulate where the curvature of the conductor is large (we will prove it this below)

The Proof:

Consider the figure above. A small conducting sphere of radius R1 and  a larger conducting sphere of radius R2. They are connected by a wire, which will ensure the potential of both sphere are the same. Since they are conducting sphere, the charges will be spread out evenly on the surface. We will assume the charge density \(\sigma =Q/(4\pi R^2) \)

The potential of a sphere at the surface is given by: 

$$ \phi=kQ/R, \quad Q=\sigma\times 4\pi R^2 $$ substituting for Q, we get:

$$\phi=4\pi k\sigma R $$

Applying to both conducting spheres and that the potential of the two are same due to wire connecting it:

$$ \phi =4\pi k\sigma_1 R_1 = 4\pi k\sigma_2 R_2 $$  

or  $$\sigma_1 R_1=\sigma_2 R_2$$  

Therefore for small \(R_1 \)  the charge density,  \( \sigma_1 \),  has to be higher. The smaller sphere has high curvature. 

Hence we prove that higher curvature means the charge density will be higher.

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Bockenheimer Warte - German train station that motivates public to Science & Knowledge

In Frankfurt, there is a train station - Bockenheimer Warte that promotes science!

I wish every country has few train station like this one, that motivates young minds to become science teacher, scientist, engineers. This world puts too much emphasis to the movie stars, glamor, religion, and sport heros. The real fact is, the teachers, scientists and engineers are the real heros. It needs to be emphasised.

If you are a reader that is connected to politicians, please help your country by encouraging these true heros. 

Here are the videos and pictures:

  

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Guitar - Easiest song to Play - Happy Birthday Song

Easiest Song to Play

Skills needed: Three chords G, C, D. Able to transition quickly. No strum skill needed. No timing skill needed.

I got this information from youtube "10 Easy Songs 3 Easy Chords G C D" from Lee John Blackmore ( at time 11:29). Click here to directly reach to that song:

 https://youtu.be/zi6CRi0SXLM?si=VCA6gXjOLhAReopb&t=11m29s

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