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Explain the meaning of the statement ‘electric charge of a body is quantised'.

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Question Id:  PH0000008 Question: Explain the meaning of the statement ‘electric charge of a body is quantised'. Answer : Charge( may be positive or negative) of any body is integral multiple of electronic charge. This phenomena is quantisation of charge. If a body has \(Q\) amount of charge then it can be say from the quantisation of charge that \(Q=\pm ne\) where \(e\) is electronic charge [ \(e=-1.6 \times 10^{-19}C\) ] and \(n=1,2,3,4,......\) Fig-1: Quantisation of charge Explanation : A body get charged in two way By losing some electron By gaining some electrons An atom has equal number of proton and electron due to ths fact in general an atom is neutral but w hen a body lose some electron then it become positively charged in this case number of proton is more in the atom and when gain some electron it become negatively charged as in this case number of electron is more in the atom. As fraction number of electron can not be transferred (lose or gain) so it means a bod...

Four point charges qA=2μC, qB=−5μC, qC=2μC, and qD=−5μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?

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Question Id:  PH0000007 Question :   Four point charges qA = 2 µC, qB = –5 µC, qC = 2 µC, and qD = –5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square? Answer : zero  Explanation :  According to the given question you can draw the figure look like as given below.   Fig-1: Coulomb's Force at the centre of square Let, side of the square is a unit. Then length of the corner is \(a\sqrt{2}\) . So, the distance froom centre to corner is \(\frac{a}{\sqrt{2}}\) . Given \(2\mu C\) charge will apply foce towards the centre. For both the \(2\mu C\) the applied forces on \(1\mu C\) chrge are equal and opposite to each other. And value of the force is  \(F=\frac{1}{4 \pi \epsilon_0} \times \frac{1 \times 2}{(\frac{a}{\sqrt 2})^{2}} \mu N\) \[\Rightarrow F=\frac{1}{4 \pi \epsilon_0} \times \frac{4}{a^{2}} \mu N\] As the force are opposite and equal so  the resultant...

The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

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Question Id:  PH0000006   Question :  The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Answer : (a) Distance between the two spheres is 0.12m. (b) Force on the second sphere due to the first sphere is save and i.e  0.2 N. Explanation : Given,  two small charge \(q_{1}=0.4\mu C=0.4\times 10^{-6}C=4\times 10^{-7}C\) ,  \(q_{2}=-0.8\mu C=-0.8\times 10^{-6}C=-8\times 10^{-7}C\)   and force(F)  between them is 0.2 N. Distance between the sphere of charges is r (say) Figure-1: force between 0.4 µC and –0.8 µC According to Coulomb's law  If two point charge \(q\) and \(Q\) are separated by a distance \(r\) then electrostatic force acting on them is \(F=\frac{1}{4\pi \epsilon }\frac{qQ}{r^{2}}\)     ;[where \(\epsilon\) is electric permittivity ...

What is the force between two small charged spheres having charges of 2×10−7C and 3×10−7C placed 30cm apart in air ?

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Question Id:  PH0000005 Question :  What is the force between two small charged spheres having charges of \(2\times 10^{-7}C\) and \(3\times 10^{-7}C\) placed 30 cm apart in air? Answer :   F orce acting  between \(q\) and \(Q\) is \(6\times 10^{-4}N\) Explanation :   Given two charge \(q=2\times 10^{-7}C\) and  \(Q=3\times 10^{-7}C\) And they are separated by a distance \(r=30cm \Rightarrow r=0.3m\) Figure-1 According to Coulomb's law  If two point charge \(q\) and \(Q\) are separated by a distance \(r\) then electrostatic force acting on them is \(F=\frac{1}{4\pi \epsilon }\frac{qQ}{r^{2}}\)     ;[where \(\epsilon\) is electric permittivity and  \(\epsilon =k\epsilon_{0}\) where k is dielectric constant and  \(\epsilon_{0}\) is electric permittivity for air medium] Now Force between \(q\) and \(Q\) is  \(F=\frac{1}{4\pi \epsilon _{0}}\frac{qQ}{r^{2}}\) ;[at air medium \(\epsilon =\epsilon_{0}\)  as \(k=1\)] \[\Right...

Express \(Sine(sin)\), \(Cosine(cos)\) and \(tangent(tan)\) in terms of sides of triangle

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Question : Express \(Sine(sin)\), \(Cosine(cos)\) and \(tangent(tan)\) in terms of sides of triangle  Answer : For a right angle triangle  we can find sine or sin, cosine or cos, tangent or tan in terms of their side.  To know more accurately about sin, cos and tan we must have to know about (1)base, (2)height (perpendicular) and (3)hypotenuse. Figure-1 (right angle triangle)    In general we use base, hypotenuse and height(perpendicular) as shown in figure.  But this method is not   always right.    Let's know what is right (1) Definition of hypotenuse:       The side opposite to the right angle or \(90^{o}\) is called hypotenuse, it is the largest side of right angle triangle.  (2) Definition of perpendicular:   It depends on the angle which we are going to consider. If we consider angle C the  perpendicular  is AB.  If we consider angle A then  perpendicular  is  BC. (See be...

A right angle triangle has a base of \(25m\) and its area is \(375m^{2}\) find its height

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Question : A triangle has a base of  \(25m\)  and its area is  \(375m^{2}\)  find its height?  Answer : Height of the right angle triangle is  \(30m\)  . Explanation : Given base is  \(25m\)  and area is  \(375m^{2}\) Figure-1 (right angle triangle) Now we know area of right angle triangle=  \(\frac{1}{2}\times base \times height\) So according to question,  \(\frac{1}{2}\times base \times height\) =375 \[\Rightarrow base \times height= 375 \times 2\] \[\Rightarrow 25 \times height= 750\] \[\Rightarrow height= \frac{750}{25}\] \[\Rightarrow height= 30\] Hence, height of the right angle triangle is \(30m\)  . (Answer)

Project on atomic model and history || Class 12 physics Project on atomic model and history

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Project on atomic model and history:            Atom is the smallest constituent unit of ordinary matter that constituent of chemical elements viz electron, proton, neutron. Atoms are extremely small, typical sizes are around 100 picometers. In 1800 AD Jhon Dalton use the concept of atom to explain why chemical elements seemed to combine in ratios of small whole numbers. And Dalton give the atomic theory famous as "Dalton's atomic theory". Atom is mainly consist of two main part as nucleus and electron shell. The nucleus is made of proton and neutron. Proton is positively charged and neutron has no charge i.e neutral. And electron is negatively charge. diagram of electron and proton Discovery of Electron:      In 1897, J J Thomson  discovered that cathode rays are not electromagnetic waves but are made of particles taht are 1800 times lighter than hydrogen (the lightest atom). And that particle are electron and it is negatively charged. ...