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we take positive charge as a point charge because positive charge has a high potential and electrons are always move from lower potential to the higher potential.so , we use positive charge as a point charge


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Every time you look up "the" spherical mirror formula, it comes with a set of "where's". These define what each symbol stands for, and the sigh convention to use to distinguish the location of objects and images and the difference between concave and convex radii. You can find different-looking spherical mirror formulas, with (naturally) different sets ...


1

According to your statement "x per y per z", I read it as x/y come first then z comes later. So it's (x/y)/z. But, this can be mistaken and it's better if we can include bracket when spell an equation.


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Two examples that come to my mind are acceleration and intensity. Acceleration is measured (in SI units) in "meters per second squared" = $\text{m}/\text{s}^2$, but it is also commonly said as "meters per second per second," which matches your first option. Likewise, intensity is measured (in SI units) in "joules per square meter per second" = ...


1

If I remember well, this section of Peskin & Schroeder uses the Wick rotation to solve integrals of type $$\int \frac{1}{k^n + ...} \cdot ... $$ Where $k^n = k \cdot k \cdot k ... (\rm n \, times)$. By performing the Wick rotation we suddenly get a spherically symmetric problem which enables us to use the well known tricks for such a case. But notice ...


3

The natural choice is actually $\mathcal{L}=\text{something}$, the reason being is that the $\sqrt{-g}$ term is naturally paired with the volume form $d^4x$. Even before considering curved spacetime, consider non Cartesian coordinates. For example spherical coordinates $$ dt\,dx\,dy\,dz = r^2 \sin\theta\ dt\,dr\,d\theta\,d\phi$$ Where does that term $r^2 ...


1

If I take a mass $m$ and apply a force $F$ (greater than $mg$) to it for a distance $h$ upwards then I will do work of: $$ W = Fh \tag{1} $$ The force $F$ has to be greater than the force due to gravity, $mg$, or the object won't move upwards, so let's write the force I apply to the mass as: $$F = mg + F'$$ then equation (1) becomes: $$\begin{align} W ...


1

You might want to change your question title to "Work done by gravity," because that is what is implied by the variables mgh. Of course, you can add a greater force than that of gravity, which would cause whatever object to which the force is applied to accelerate (since the forces are not in balance). No matter how much force you apply, the force of ...


1

It depends on where you are standing. A convex surface is one that bulges out towards the person who is talking about it. Since we do not live inside glass, everyone knows what a bi-convex glass lens is - one that bulges out on both sides as seen by someone who lives in air. Note that while it bulges out towards the incident ray arriving from the air, the ...


0

How would you define a convex surface? I'd call a set $\mathcal S$ (of five or more elements) "convex" if for each element $A \in \mathcal S$ any two circumspheres of $A$ and any triples of distinct elements of $\mathcal S$ other than $A$ have the property that the center of the smaller circumsphere is inside the larger circumsphere; in other words: ...


1

In your example #1, you need to remember that mass is conserved. The stream of exhaust left behind by the rocket grows in mass by exactly the rate the rocket consumes fuel. In your example #2, there's often a desire to make constants positive. If that constant is actually negative in your interpretation of the problem, you need to add a minus sign. In your ...


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It's a matter of history. When George Stoney developed Stoney units in 1881, or when Robert Millikan performed the oil drop experiment in 1909, it wasn't yet known that it was possible for anything to have a charge smaller in magnitude than the charge of an electron. By the time the quark model was proposed, in 1964, the use of the "elementary charge" being ...


4

If the particle moves from the point $x$ to $x+dx$, and assume $dx\gt 0$ for simplicity, then its potential energy increases by $$ dU = \frac{dU}{dx}dx $$ Well, it increases if $dU$ is positive and decreases if $dU$ is negative. So far I have only used the definition of the derivative – pure mathematics. However, the total energy is conserved. The sum of ...



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