Electrostatics (Only Kidding)

January 30, 2024January 30, 2024

202401301506 Solution to 2000-CE-PHY-II-28

Two insulated uncharged metal spheres $X$ and $Y$ are placed in contact. A positively-charged rod is brought near $X$ as shown below. $X$ is then earthed momentarily. The charged rod is removed and the two spheres are then separated. Describe the charges on $X$ and $Y$ . _(modified)

Roughwork.

Have some mental pictures (*make drawings if you don’t have ample RAM).

First, when a positively-charged rod is brought near $X$ , charges are induced:

Notice that on surface areas $X$ and $Y$ making contact, as boxed below:

the net charge is zero, $Q=0$ . Hence, the plus and the minus signs (representing positive and negative charges) are erased from the drawing:

Then, $X$ is earthed momentarily:

it is convenient to treat the two spheres $X$ and $Y$ as one body $X+Y$ :

The process of earthing enables a transfer of negative free charges (electrons being the carrier) unidirectionally between two bodies:

here from the earth to the body, but no reversed (why?); the plus and minus signs, as boxed below:

should cancel off each other, and are thus erased from the drawing:

once the charged rod is removed:

the negative charges will be redistributed over $X+Y$ such that the electrostatic repulsive forces between them are kept to a minimum:

after the two spheres are separated,

both spheres $X$ and $Y$ will be negative charged. The descriptions go complete.

This problem is not to be attempted.

November 8, 2023December 10, 2023

202311081445 Pastime Exercise 006

The figure below is a representation of the $\mathbf{E}$ -field by electric field lines in a family $R$ of infinitely many quadratic functions

$y_t(x_{t'})=a_tx_{t'}^2+b_tx_{t'}+c_t$

Roughwork.

A tangent to any quadratic curve at some point in the locus gives the $\mathrm{\pm ve}$ direction of the electric force experienced by a (positive) test charge placed there. I.e.,

$\displaystyle{T(x_{t'},y_{t})=\frac{\mathrm{d}}{\mathrm{d}x}\big( y_t(x_{t'})\big) = 2a_tx_{t'}+b_{t}}$ ,

the slope of tangent.

WLOG we work with the first quadrant. First, begin with the repulsive force $\mathrm{}_Q\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $+Q$ .

$\begin{aligned} \mathrm{}_QF_{q,x}^2+\mathrm{}_QF_{q,y}^2 & = \mathrm{}_QF_{q}^2 \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg)\mathrm{}_QF_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_QF_{q,x} & = \frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}} \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg) \Bigg(\frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_Q\mathbf{F}_{q} & = \mathrm{}_QF_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_QF_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Next, continue with the attractive force $\mathrm{}_{-Q}\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $-Q$ .

$\begin{aligned} \mathrm{}_{-Q}F_{q,x}^2+\mathrm{}_{-Q}F_{q,y}^2 & = \mathrm{}_{-Q}F_{q}^2 \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg)\mathrm{}_{-Q}F_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}F_{q,x} & = \frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}} \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg) \Bigg(\frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}\mathbf{F}_{q} & = \mathrm{}_{-Q}F_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_{-Q}F_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Adding $\mathrm{}_{Q}\mathbf{F}_{q}$ and $\mathrm{}_{-Q}\mathbf{F}_{q}$ will give the resultant electric force $\mathbf{F}_{E}(x_{t'},y_t)$ .

Note that

$\begin{aligned} \mathrm{}_QF_q & = k\frac{Q}{\mathrm{}_Qr_{q}^2} \\ \mathrm{}_{-Q}F_q & = k\frac{-Q}{\mathrm{}_{-Q}r_{q}^2} \\ \end{aligned}$

where

$\begin{aligned} \mathrm{}_Qr_q & = \sqrt{(d/2+x_{t'})^2+y_t^2} \\ \mathrm{}_{-Q}r_q & = \sqrt{(d/2-x_{t'})^2+y_t^2} \\ & \\ & \\ \end{aligned}$

are the distances of a test charge at $q(x_{t'},y_t)$ from point charges $+Q$ at $A(-d/2,0)$ and $-Q$ at $B(d/2,0)$ .

The smallest angle between $\mathbf{F}_{E}(x_{t'},y_{t})$ and the level should be equal to the slope of tangent $2a_tx_{t'}+b_{t}$ at point $q(x_{t'},y_t)$ .

As of the vertex of each parabola, the $x$ -coordinate is

$\displaystyle{0=x_{t'}=-\frac{b_{t}}{2a_{t}}}$

s.t. $b_{t}=0$ , and the $y$ -coordinate $c_{t}=y_t(0)$ .

I guess, under correction, the separation distance $d$ is none any parameter of the loci.

Visualisation is $\textrm{\scriptsize{NOT}}$ mathematical $\textrm{\scriptsize{BUT}}$ conceptual.

February 21, 2019January 27, 2023

201902210410 Short Review I (Electric Charge)

Electrostatics is the study of charges at rest.

Two kinds of charge

1. There are two kinds of charge—the positive charge and the negative charge.
2. Like charges repel; unlike charges attract.
3. Since neutrons have no charge, it is said to be neutral.
4. When an object is neither positively nor negatively charged, it is also said to be (electrically) neutral.

How to charge an object?

By (a) rubbing/friction, by (b) inducing charges, or by (c) contact/sharing, there is always a transfer of free charges.

(Capacitors, made up of conductors, can be charged by Extra-High Tension (EHT) power supply.)

Where do charges come from?

Only when an object gains electrons it becomes negatively charged; only when an object loses electrons it becomes positively charged.

Conservation of charge: Charge cannot be created or destroyed.

Unit of charge

The unit is coulomb, written as $\mathrm{C}$ . The charges of a proton and of an electron are $+e=1.6\times 10^{-19}\mathrm{C}$ and $-e=-1.6\times 10^{-19}\mathrm{C}$ respectively.

For your information:

The Triboelectric Series (shortened)

It is the relative position of two materials in this series that determines which one receives electrons and which one donates. The one nearer to the positive end will be positively charged when being rubbed with the other less near, which will then be negatively charged.

$(-)$ Negative End of Series: Silicon rubber $<$ Teflon $<$ Polyethylene $<$ Saran $<$ Orlon $<$ Synthetic rubber $<$ Brass and silver $<$ Nickel and copper $<$ Hard rubber $<$ Sealing wax $<$ Amber $<$ Wood $<$ Steel $<$ Cotton $<$ Paper $<$ Aluminium $<$ Silk $<$ Lead $<$ Wool $<$ Nylon $<$ Glass $<$ Acetate $<$ Asbestos $<$ Human Hands : $(+)$ Positive End of Series

Concept Test

Which of the following statements is correct?
1. Given that glass rod is positively charged after it is rubbed with a piece of silk, it is also positively charged after it is rubbed with any other materials.
2. An object could be negatively charged by transferring away some of its protons.
3. After rubbing two different materials together, each material is charged with the same quantity of charge but with opposite sign.
4. In the process of inducing charges, an object’s net charge either increases or decreases.
Which of the following statements is/are correct?
1. 1. Each carbonate ion, $\mathrm{CO_3\,^{2-}}$ , has more electrons than protons.
  2. Sodium ion $\mathrm{Na^{+}}$ has one positive coulomb of charge.
  3. Chloride ion $\mathrm{Cl^{-}}$ has $-1.6\times 10^{-19}\mathrm{C}$ of charge.

1. I only.
2. I and II only.
3. I and III only.
4. All of the above.

Answers:

Explanation:

A is wrong: for instance, the glass rod would become negatively charged when it is rubbed with asbestos. Please refer to The Triboelectric Series. B is wrong because the protons are always bound with the neutrons in the nucleus. Only free electrons could be transferred. C is correct because only as such could the charge be conserved. D is wrong because net charge of an object must be unchanged, unless earthing occurs, which will be discussed later on.
I is correct. The superscript $(2-)$ indicates it has two extra electrons than protons. II is wrong, for the superscript $(+)$ means one extra proton than electron so that the charge should be $+e$ , or $+1.6\times 10^{-19}\mathrm{C}$ . By the same reason, III is correct.

Conductors and Insulators

Conductors allow free charges (e.g., electrons, electrolytes) to flow through them easily; insulators do not.

(Not all conductors are metal, e.g., electrolyte is not a metal but a conductor.)

An isolated and charged conductor always has its charges distributed on its surface; the charge density is higher on curved edges and cusps than on flat and smooth surface.

Friction_Induce_Contact_Earthing_Illustration

The above chart illustrates the three processes of charging and the process of earthing. In doing experiments, and also exercises, we often use a combination of some of these four techniques, so make sure you understand and remember them.

Coulomb’s Law

$F=k\times \displaystyle{\frac{Q_1Q_2}{r^2}}$

where $k=\displaystyle{\frac{1}{4\pi \epsilon_0}}$ is the proportionality constant/Coulomb constant.

$F = \displaystyle{\frac{Q_1Q_2}{4\pi \epsilon_0r^2}}$

where $\epsilon_0=8.85\times 10^{-12}\mathrm{C^2N^{-1}m^{-2}}$ is the permittivity of free space.

The magnitude of Coulomb force is given above. The direction of the force on each point charge is pointing away from each other when the two point charges $Q_1$ and $Q_2$ are of the same sign, whereas pointing towards each other when of opposite sign.

(Point charge, as well as point mass, is an ideal model of particle. When the size of two charge carriers is exceedingly less than their separating distance, we treat them as point charges. For example, the radius of an electron is approximated to $10^{-15}$ order of magnitude, much less than the radius of an atom $10^{-10}$ .)

Concept Test

1. A positively charged glass rod is brought near (without contact) an isolated aluminium rod which is electrically neutral.
  
  Which of the following statements is correct?
  1. The glass rod induces a net negative charge on the aluminium rod.
  2. The electrons are transferred from the aluminium rod to the glass rod.
  3. The net electrostatic force acting on the aluminium rod is zero.
  4. The net electrostatic force acting on the aluminium rod is non-zero.
2. Sally wants to charge a metallic conductor by friction. After rubbing it with a tablecloth for 5 minutes, she still finds the conductor uncharged. Which of the following statements could best explain the failure?
  1. Conductors can never be charged by friction.
  2. Sally shall continue the rubbing for a longer time.
  3. The charge obtained by the conductor during rubbing flows away through Sally’s body.
  4. The air surrounding the rubbed surface of the conductor is ionized such that the charges acquired flow away through air.
3. Two identical spherical conductors of radius are separated by a distance of between their origins. Each of them contains the same amount of charges . But one is positively charged; another is negatively charged. Which of the following about electrostatic force is true?
  1. $F=\displaystyle{k\frac{(Q)(-Q)}{(5r)^2}}$
  2. $F=\displaystyle{k\frac{(Q)(-Q)}{(3r)^2}}$
  3. None of the above.

Answers:

Explanation:

Unless there is earthing on the aluminium rod, its net charge is conserved. A is wrong. Unless being in contact, the electrons are never transferred from the aluminium rod to the glass rod in mid-air. B is wrong. This aluminium rod would be induced negative (opposite) charges on its front side near the glass rod and positive (like) charges on the back side away from the glass rod. Because the aluminium’s front side is closer to the glass rod than its back side, and electrostatic force $\displaystyle{\varpropto \frac{1}{r^2}}$ , we know the force of attraction is greater than that of repulsion. So there is a non-zero net force acting on the aluminium rod. C is wrong and D is correct.
When Sally’s hand touches the metallic conductor, the conductor is earthed. So the charge acquired by friction is discharged first to her body and then to the ground. Thus, to charge a metallic conductor by friction, it must be held with an insulator.
The Coulomb’s force law does not apply here because we cannot treat the two spherical conductors as point charges, for two reasons: (i) the charge distribution of both conductors is uneven—the charges are denser at the surfaces close to each other, and (ii) the separating distance $5r$ is only five times the radius $r$ of the spherical conductors, they are in same order of magnitude.