## विद्युत क्षेत्र

विद्युत क्षेत्र

प्रत्येक आवेश अपने चारों ओर एक निश्चित क्षेत्र उत्पत्र करता है। यदि एक आवेश $\displaystyle {{Q}_{1}}$ इसके नजदीक स्थित एक अन्य आवेश $\displaystyle {{Q}_{2}}$ पर बल आरोपित करता है तो यह कहा जा सकता है कि $\displaystyle {{Q}_{2}}$$\displaystyle {{Q}_{1}}$ के क्षेत्र में रखा हुआ है इसलिए यह बल का अनुभव करता है या
Q1, Q2 के क्षेत्र में रखा है इसलिए यह बल का अनुभव करता है इस प्रकार एक आवेश के चारों ओर वह क्षेत्र, जिसमें कोई अन्य आवेश बल का अनुभव करता है, विद्युत क्षेत्र कहलाता है।

विद्युत क्षेत्र की तीव्रता $\displaystyle (\vec{E})$: विद्युत क्षेत्र में किसी बिन्दु पर स्थित एकांक धन_आवेश जितने बल का अनुभव करता है उसे उस बिन्दु पर विद्युत क्षेत्र की तीव्रता कहते हैं। $\displaystyle \vec{E}= \frac{{\vec{F}}}{{{q}_{0}}}$

जहाँ $\displaystyle {{q}_{0}}\to 0$ ताकि इस आवेश की उपस्थिति आवेश स्त्रोत Q को प्रभावित न करे और इसके विद्युत क्षेत्र में कोई परिर्वतन न हो इसलिए विद्युत क्षेत्र की तीव्रता के लिए सही व्यंजक इस प्रकार है $\displaystyle \vec{E}=\underset{{{q}_{0}}\to 0}{\mathop{Lim}}\,\,\,\,\frac{{\vec{F}}}{{{q}_{0}}}$

मात्रक एवं विमीय सूत्र : SI मात्रक एवं CGS मात्रक

विमाएँ [$\displaystyle E$] =[$\displaystyle ML{{T}^{-3}}{{A}^{-1}}$]

विद्युत क्षेत्र की दिशा : विद्युत क्षेत्र (तीव्रता){ $\displaystyle \vec{E}$ एक सदिश राशि है। धनावेश के कारण विद्युत क्षेत्र की दिशा सदैव आवेश से दूर की ओर एवं रुणावेश के कारण आवेश की ओर होती है।

विद्युत बल एवं विद्युत क्षेत्र में सम्बन्ध : विद्युत क्षेत्र $\displaystyle \vec{E}$ में रखा एक आवेश ; (Q), $\displaystyle F=QE$ बल का अनुभव करता है। यदि आवेश धनात्मक है तो बल की दिशा विद्युत क्षेत्र की दिशा में होती है और यदि रुणात्मक है तो बल की दिशा विद्युत क्षेत्र की विपरीत दिशा में होती है।

विद्युत क्षेत्र का अध्यारोपण (कई आवेशों के कारण किसी एक बिन्दु पर विद्युत क्षेत्र): कई आवेशों के कारण किसी एक बिन्दु पर परिणामी विद्युत क्षेत्र, प्रत्येक आवेश के कारण उस बिन्दु पर उत्पत्र विद्युत क्षेत्र के सदिश योग के तुल्य होता है।

$\displaystyle \vec{E}={{\vec{E}}_{1}}+{{\vec{E}}_{2}}+{{\vec{E}}_{3}}+...$

दो विद्युत क्षेत्रों के परिणामी विद्युत क्षेत्र का परिमाण

$\displaystyle E=\sqrt{E_{1}^{2}+E_{2}^{2}+2{{E}_{1}}{{E}_{2}}\cos \theta }$ एवं दिशा $\displaystyle \tan \alpha =\frac{{{E}_{2}}\sin \theta }{{{E}_{1}}+{{E}_{2}}\cos \theta }$

सतत एवं एकसमान आवेश वितरण के कारण विद्युत क्षेत्र : इस प्रकार के वितरण में कई आवेश एक निश्चित अल्प दूरी पर व्यवस्थित होते हैं।

## विद्युतदर्शी (Electroscope) क्या है

विद्युतदर्शी

यह एक सरल उपकरण है, जिसकी सहायता से किसी वस्तु पर आवेश की उपस्थिति को ज्ञात किया जाता है। जब एक आवेशित वस्तु को इसकी धात्विक घुण्डी (Knob) के सम्पर्क में लाते हैं तो कुछ आवेश स्वर्ण पतियों (Leaves) पर स्थानान्तरण हो जाता है, एवं प्रतिकर्षण के कारण ये पतियाँ फैल जाती हैं। यदि एक आवेशित वस्तु को पहले से आवेशित विद्युतदर्शी के नजदीक लाते हैं और यदि वस्तु का आवेश और विद्युतदर्शी पर उपस्थित आवेश एक ही प्रकृति का है तो पतियाँ ओर अधिक फैल जाती हैं और यदि विपरीत प्रकृति का है तो पतियाँ सामान्यत: सिकुड़ (Converge) जाती है परन्तु यदि प्रेरण प्रभाव बहुत तीव्र है तो पतियाँ सिकुड़ने के बाद फिर से फैल जाती हैं।

अनावेशित विद्युतदर्शी

आवेशित विद्युतदर्शी

## Real Numbers – Revisiting Rational and Their Decimal Expansions

REVISITING RATIONAL NUMBERS AND THEIR DECIMAL EXPANSIONS :

We have already studied in the previous class that rational numbers have either a terminating decimal repeating decimal expansion. We have to consider a rational number as $\displaystyle \frac{p}{q}$ (where $\displaystyle q\ne 0$) as terminating or non-terminating repeating (or recurring) decimal expansion.

e.g.

(A) (i)    0.0527    =

(ii)    26.12489     =

(B) (i)    0.0875    =

(ii)    23.3408    =

Now, we have converted a real number whose decimal expansion terminates into a rational number of the form. $\displaystyle \frac{p}{q}$, where p and q are coprime, and prime factorization of denominator

(i.e., q) has only power of 2 or power of 5 or both. We should also understand that the denominator is a power of 10 and can be only prime factors 2 and 5.

Now, we get that this real number is a rational number of the form $\displaystyle \frac{p}{q}$, where the prime factorization of q is of the form 2n, 5m, and n, m are some non-negative integers.

Theorem : Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form $\displaystyle \frac{p}{q}$, where p and q are coprime, and the prime factorization of q is of the form 2n5m, where n, m are non-negative integers.

Let $\displaystyle \frac{a}{b}$ be a rational number in the lowest form such that the prime factorization of b is of the form $\displaystyle {{2}^{n}}\times {{5}^{m}}$ . Where n and m are non-negative integers.

Example:

(i) $\displaystyle \frac{3}{8}=\frac{3}{{{2}^{3}}}=\frac{3\times {{5}^{3}}}{{{2}^{3}}\times {{5}^{3}}}=\frac{375}{{{10}^{3}}}=0.375$

(ii) $\displaystyle \frac{13}{125}=\frac{13}{{{5}^{3}}}=\frac{13\times {{2}^{3}}}{{{5}^{3}}\times {{2}^{3}}}=\frac{104}{{{10}^{3}}}=0.104$

(iii) $\displaystyle \frac{2139}{1250}=\frac{2139}{2\times {{5}^{4}}}=\frac{2139\times {{2}^{3}}}{{{2}^{4}}\times {{5}^{4}}}=\frac{2139\times 8}{{{\left( 2\times 5 \right)}^{4}}}=\frac{17112}{{{10}^{4}}}=1.7112$

Theorem : Let $\displaystyle x=\frac{p}{q}$ be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Example:

$\displaystyle \frac{1}{7}$

In this case, denominator 7 is not of the form 2n5m. So, from above mentioned theorems, $\displaystyle \frac{1}{7}$ will not have a terminating decimal expansion. Hence, 0 (zero) will not be remainder and the remainders will start repeating after a certain stage. We have a block of digits, 142857, repeating in the quotient of $\displaystyle \frac{1}{7}$.

Hence, it is is true that any rational number is not converted by the above theorems.

Theorem :Let $\displaystyle x=\frac{p}{q}$ be a rational number, such that the prime factorization of q is not of the form 2n5m, where n, m are non-negative integers. Then, x has a decimal expansion which is not-terminating repeating (recurring).

Hence, we can say that the decimal expansion of every rational number is either terminating or non-terminating repeating (recurring).

## Real Number-Revisiting Irrational Numbers

REVISITING IRRATIONAL NUMBERS :

We have studied in the previous class about rational numbers as well as irrational numbers. Both are the members of the REAL NUMBERS family. We have also studied how to locate irrational numbers on the number line.

Any rational number is represented by $\displaystyle \frac{p}{q}$; where $\displaystyle q\ne 0$ whereas irrational number is represented by ‘s’

0.10110111011110….. etc. are the examples of irrational numbers.

Theorem : Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.

Proof :

Let the prime factorization of a be as follows :

a = p1, p2, p3, p4, p5, …., …. pn

$\displaystyle \therefore$ a2 = (p1, p2, p3, p4, …. Pn) (p1, p2, p3, p4, …. Pn)

= p12, p22, p32, p42, …………… pn2.

It is given that p divides a2, so from the Fundamental Theorem of Arithmetic p is one of the prime factors of a2. Also a2 having prime factors p1, p2, p3, ….. pn.

Hence, p is one of p1, p2, p3, …… pn.

Example 6: To prove $\displaystyle \sqrt{2}$ is irrational.

Solution. Let us assume that $\displaystyle \sqrt{2}$ is rational.

Now, we can find integers r and $\displaystyle s\left( \ne 0 \right)$ such that $\displaystyle \sqrt{2}=\frac{r}{s}$.

Again let r and s have a common factor other than 1. Then we can divide by common factor to get, where a and b are coprime.

$\displaystyle \Rightarrow$  $\displaystyle b\sqrt{2}=a$ …….. (i)

$\displaystyle \Rightarrow$  $\displaystyle {{\left( b\sqrt{2} \right)}^{2}}={{\left( a \right)}^{2}}$ [squaring both sides]

$\displaystyle \Rightarrow$ 2b2 = a2

$\displaystyle \therefore$ 2 divides a2.

Now, by above theorem

2 divides a also.

So, we can write a = 2c for some integer c.

$\displaystyle \Rightarrow$  $\displaystyle b\sqrt{2}=2c$ [From (i)]

$\displaystyle \Rightarrow$ 2b2 = 4c2 [Squaring both sides]

$\displaystyle \therefore$ b2 = 2c2 ……. (ii)

i.e., 2 divides b2

So, 2 divides b [by using above theorem with p = 2]

Again, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that $\displaystyle \sqrt{2}$ is rational.

Hence, we conclude that $\displaystyle \sqrt{2}$ is irrational.

## Real Numbers-The Fundamental Theorem of Arithmetic

THE FUNDAMENTAL THEOREM OF ARITHMETIC

We have already studied in the previous classes that any natural can be written as a product of its prime factors. e.g. = 3 = 3, 6 = 2 $\displaystyle \times$ 3, 275 = 11 $\displaystyle \times$ 25 etc. i.e, any natural number can be obtained by multiplying prime numbers.

If we take prime numbers like 2,3,5,7,11,13, and 29 and if we multiply some or all of these numbers, allowing them to repeat as we can, then we can get a large collection of positive integers (infinitely).

e.g.             11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 4147

7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 29029

5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 145145

3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 435435

2 $\displaystyle \times$3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 870870

22 $\displaystyle \times$3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 1741740

23 $\displaystyle \times$3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 3483480

Now, we see there are infinitely many primes and so we can get an infinite collection of numbers by all the primes and all possible products of primes.

In the other hand, if we factorise positive integers, we have to do the opposite of what we have done so far.

We can use factor tree for factorise of 65520 as shown below.

Hence, we have factorised 65520 as 2 $\displaystyle \times$ 2 $\displaystyle \times$ 2 $\displaystyle \times$2 $\displaystyle \times$ 3 $\displaystyle \times$ 3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 13 as a product of primes, i.e., 65520 = 24 $\displaystyle \times$ 32 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 13 as a product of powers of primes.

If we take 123456789 then it is equal to 32 $\displaystyle \times$ 3803 $\displaystyle \times$ 3607.

In these factors 3803 and 3607 are primes. It means we have to understand that every composite number can be written as the product of primes. It is called Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers.

Hence, we can say Fundamental Theorem of Arithmetic as follows :

Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur, i.e., for any composite number there will be one and only one way to express it as a product of primes, as long as we are not particular about the order in which the primes occur.

e.g    2 $\displaystyle \times$ 3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 as the same as 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 2 $\displaystyle \times$ 3 or any other possible order in which these primes are written. It can be defined as follows :

The prime factorization of a natural number is unique, except for the order of its factors.

If any composite number is x, then the factor of x = p1 , p2 , p3 , …. Pn, where p1, p2, p3, …. Pn are primes and written in ascending order, i.e., $\displaystyle {{p}_{1}}\le {{p}_{2}}\le {{p}_{3}}......\le {{p}_{n}}$.

If we combine the same primes, we will get powers of primes.

E.g.    65520 = 2 $\displaystyle \times$ 2 $\displaystyle \times$ 2 $\displaystyle \times$ 2 $\displaystyle \times$ 3 $\displaystyle \times$ 3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$13

24 $\displaystyle \times$ 32 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$13

When we decide that the order will be ascending, then the way the number is factorised, is unique.

There are many applications of the Fundamental Theorem of Arithmetic in mathematics as well as in other fields.

Example 4:Consider the number 16n, where n is a natural number. Check whether there is any value of n for which 16n ends with the digit zero.

Solution. If the number 16n, for any n, were to end with the digit zero, then it would be divisible by 5 and so its prime factorization must contain the prime 5.

16n = (24)n = 24n

The only prime in the factorization of 16n is 2.

There is no other primes in the factorization of 16n = 24n

[By uniqueness of the Fundamental Theorem of Arithmetic]

5 does not occur in the prime factorization of 16n for any n.

4n does not end with the digit zero for any natural number n.

We have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in previous classes, without realising it. This method is also called the Prime factorization method.

Example :

Find the LCM and HCF of 18 and 60 by the prime factorization method.

Solution.

18 = 2 $\displaystyle \times$ 3 $\displaystyle \times$ 3 = 21 $\displaystyle \times$ 32

60 = 2 $\displaystyle \times$ 2 $\displaystyle \times$ 3 $\displaystyle \times$ 5 = 22 $\displaystyle \times$ 31 $\displaystyle \times$ 51

We get HCF of 18 and 60 as 2

And LCM of 18 and 60 as 180

But now, we have to understand that

HCF of 18 and 60 is 21 = Product of the smallest power of each common prime factor in the numbers.

LCM of 18 and 60 is 22 $\displaystyle \times$ 32 $\displaystyle \times$5 = Product of the greatest power of each Prime factor, involved in the numbers.

For any two positive integers a and b , (HCF of a and b) $\displaystyle \times$ (LCM of a and b) = a x b

## Real Numbers: Euclid’s Division Lemma

EUCLID’S DIVISION LEMMA :

Euclid was the first Greek Mathematician who gave a new way of thinking the study of geometry. He also made important contributions to the number theory. Euclid’s Lemma is one of them. It is a proven statement which is used to prove other statements.

Let ‘a’ and ‘b’ be any two positive integers. Then, there exist unique integers q and r such that

;

Now, we say ‘a’ as dividend, ‘b’ as divisor, ‘q’ as quotient and ‘r’ as remainder.

Dividend = (divisor $\displaystyle \times$ quotient) + remainder

Example: Let 578 be divided by 16.

36 is as quotient and 2 as remainder.

In this case :

Dividend = 578

Divisor = 16

Quotient = 36

And remainder = 2

Dividend = (quotient $\displaystyle \times$ divisor) + remainder

Hence, 578 = (36 $\displaystyle \times$ 16) + 2

To get the HCF of two positive integers, let ‘c’ and ‘d’, with c > d, follow as :

(i) Applying Euclid’s Lemma,

$\displaystyle c=dq+r;\,0\le r

(ii) If r = 0, d is the HCF of c and d.

If $\displaystyle r\ne 0,$ then applying the division lemma to d and r.

(iii) We can continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

This algorithm works because HCF (c,d) = HCF (d,r)

Where the symbol HCF (c,d) denotes the HCF of c and d etc.

Example: Use Euclid’s algorithm to find HCF of 425 and 40

Solution:  Let a = 425 and b = 40

By Euclid’s division lemma; we have

$\displaystyle 425=40\times 10+25$ …. (i)

By using the above theorem, we observe that the common divisors of a = 425 and b = 40 are also the common divisors of b = 40 and r1 = 25 and vice-versa.

Applying Euclid’s division lemma on divisor b = 40 and remainder r1 = 25,

We get $\displaystyle 40=25\times 1+15$ …. (ii)

$\displaystyle b=q_{2}^{r},+{{r}_{2}}$; where q2 = 1 and r2 = 15

Again using the above theorem, we find that, the common divisors of r1 = 25 and r2 = 15 are the common divisors of b = 40 and r1 = 25 and vice-versa. But the common divisors of b = 40 and r1 = 25 are the common divisors of a = 425 and b = 40 and vice-versa.

Applying Euclid’s division lemma on r1 = 25 and r2 = 15, we get

$\displaystyle 25=15\times 1+10$ …. (iii)

$\displaystyle \Rightarrow$ r1 = r2q3 + r3, where q3 = 1 and r3 = 10

Again by using the above theorem, we find that common divisors of r2 = 15 and r3 = 10 are the common divisors of a = 425 and b = 40 and vice-versa.

Now, Using Euclid’s division lemma on r2 = 15 and r3 = 10, we get

$\displaystyle 15=10\times 1+5$ …. (iv)

$\displaystyle \Rightarrow$  $\displaystyle r{{ & }_{2}}={{r}_{3}}\times {{q}_{4}}+{{r}_{4}}$; where q4 = 1 and r4 = 5

Again by using the above theorem, we find that, the common divisors of r2 = 15 and r3 = 10 are the common divisors of a = 425 and b = 40 and vice-versa.

Using Euclid’s division lemma on r3 = 10 and r4 = 5, we get

$\displaystyle 15=5\times 3+0$ …. (v)

Hence, r4 = 5 is a divisor of r3 = 10 and r4 = 5. Also, it is the greatest common divisor (or HCF) of r3 and r4. So, r4 = 5 is the greatest common divisor (or HCF) of a = 425 and b = 40. We also observe that r4 = 5 is the last non-zero remainder in the above process of repeated application of Euclid’s division lemma on the divisor and the remainder in the next step.

The set of equation (i) to (v) is called Euclid’s division algorithm for 425 and 40. The last divisor, or the last but non-zero remainder 4 is the HCF (or GCD) of 425 and 40.

The above process of finding HCF can also be carried out by successive divisions as follows:

(i) Euclid’s division lemma and algorithm are so closely interlinked that it is often called division algorithm.

(ii) Euclid’s Division Algorithm is stated for only positive integers except zero. i.e., $\displaystyle b\ne 0$.

Example 3:

A sweet seller has 840 kaju barfis and 260 badam barfis. He wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of barfis that can be placed in each stack for this purpose?

Solution. The area of the tray that is used up will be the least. For this, we find HCF (840, 260). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least.

Now, applying Euclid’s algorithm to find their HCF, we have

840    =    260 $\displaystyle \times$ 3 + 60

260    =    60 $\displaystyle \times$ 4 + 20

60    =    20 $\displaystyle \times$ 3 + 0

Hence, HCF of 840 and 260 is 20.

So, the sweet seller can make stacks of 10 for both kinds of barfi

## Real Numbers

• Introduction
• Euclid’s Division Lemma
• The Fundamental Theorem of Arithmetic
• Revisiting Irrational Numbers
• Revisiting Rational Numbers and Their Decimal Expansions
• Summary

INTRODUCTION :

We have already studied about irrational numbers in 9th class. Now, we will study the real numbers and also about the important properties of positive integers for Euclid’s division algorithm and the Fundamental Theorem of Arithmetic.

Euclid’s division algorithm says us about divisibility of integers. It states that any positive integer ‘a’ can be divided by any other positive integer ‘b’ in such a way that it has ‘r’ remainder which is smaller than ‘b’. In fact it is a long division method. We also use it to compute the HCF of two positive integers.

The Fundamental Theorem of Arithmetic says us about the expression of positive integers as the product of prime integers. It states that every positive integer is either prime or a product of powers of prime integers. We have already known how to find HCF and LCM of positive integers by using the Fundamental Theorem of Arithmetic in previous class. Now, we will learn about irrationality of many numbers like etc. by applying this theorem. We know that the decimal representation of a rational number is either terminating or if repeating if not terminating. We have to use the Fundamental Theorem of Arithmetic to determine the nature of the decimal expansion of rational numbers.